VERSION 0.9 NCS
GRADE 12
MATHEMATICS
WRITTEN BY VOLUNTEERS
SlYAVULA
TECHNOLOGYPOWERED LEARNING
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Everything Maths
Grade 12 Mathematics
Version 0.9 NCS
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Mudau;
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Mark Horner; Samuel Halliday; Sarah BIyth; Rory Adams; Spencer Wheaton
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Everything Maths
Mathematics is commonly thought of as being about numbers but mathematics is actually a language!
Mathematics is the language that nature speaks to us in. As we learn to understand and speak this lan
guage, we can discover many of nature's secrets. Just as understanding someone's language is necessary
to learn more about them, mathematics is required to learn about all aspects of the world  whether it
is physical sciences, life sciences or even finance and economics.
The great writers and poets of the world have the ability to draw on words and put them together in ways
that can tell beautiful or inspiring stories. In a similar way, one can draw on mathematics to explain and
create new things. Many of the modern technologies that have enriched our lives are greatly dependent
on mathematics. DVDs, Google searches, bank cards with PIN numbers are just some examples. And
just as words were not created specifically to tell a story but their existence enabled stories to be told, so
the mathematics used to create these technologies was not developed for its own sake, but was available
to be drawn on when the time for its application was right.
There is in fact not an area of life that is not affected by mathematics. Many of the most sought after
careers depend on the use of mathematics. Civil engineers use mathematics to determine how to best
design new structures; economists use mathematics to describe and predict how the economy will react
to certain changes; investors use mathematics to price certain types of shares or calculate how risky
particular investments are; software developers use mathematics for many of the algorithms (such as
Google searches and data security) that make programmes useful.
But, even in our daily lives mathematics is everywhere  in our use of distance, time and money.
Mathematics is even present in art, design and music as it informs proportions and musical tones. The
greater our ability to understand mathematics, the greater our ability to appreciate beauty and everything
in nature. Far from being just a cold and abstract discipline, mathematics embodies logic, symmetry,
harmony and technological progress. More than any other language, mathematics is everywhere and
universal in its application.
See introductory video by Dr. Mark Horner: VMiwd at www.everythingmaths.co.za
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9:41 AM
GRADE 10
MATHEMATICS
(2) Help : 006d
Factorise : [2k 2 j + 24fc 2 j 2
• Is 12, k and j common factors? 
• Is it best to use k2j or kj2 as [
common factor? {
• What does factorise ean?
mathImat.cs ^^^ 06 ^
Factorise : 12k 2 j + 2\k 2 f
Shud I devide out both k
and j or just one?
n
P You should take out all the
t  common factors, so 12, k 2 and
j because they appear in both
terms
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Contents
1 Introduction to Book
1.1 The Language of Mathematics
2.6 Logarithm Law 2
2.7 Logarithm Law 3
2 Logarithms 4
2.1 Introduction 4
2.2 Definition of Logarithms 4
2.3 Logarithm Bases 5
2.4 Laws of Logarithms 6
2.5 Logarithm Law 1 : log a 1 = 6
log» = 1 7
l°ga<>2/) = log a (x) +log a (j/) 8
2.8 Logarithm Law 4: log a (J = log a (x)  log a (y) 8
2.9 Logarithm Law 5: log a (x f ') = 61og a (x) 9
2.10 Logarithm Law 6: log a (^x) = ^M 10
2.11 Solving Simple Log Equations 12
2.12 Logarithmic Applications in the Real World 15
3 Sequences and Series 19
3.1 Introduction 19
3.2 Arithmetic Sequences 19
3.3 Geometric Sequences 21
3.4 Recursive Formulae for Sequences 25
3.5 Series
26
3.6 Finite Arithmetic Series 28
3.7 Finite Squared Series 31
3.8 Finite Geometric Series 32
3.9 Infinite Series 34
4 Finance 39
4.1 Introduction 39
4.2 Finding the Length of the Investment or Loan 39
4.3 Series of Payments 40
4.4 Investments and Loans 48
4.5 Formula Sheet 53
5 Factorising Cubic Polynomials 56
5.1 Introduction 56
5.2 The Factor Theorem 56
5.3 Factorisation of Cubic Polynomials 58
5.4 Solving Cubic Equations 61
11
CONTENTS CONTENTS
6 Functions and Graphs 65
6.1 Introduction 65
6.2 Definition of a Function 65
6.3 Notation Used for Functions 66
6.4 Graphs of Inverse Functions 67
7 Differential Calculus 75
7.1 Introduction 75
7.2 Limits 75
7.3 Differentiation from First Principles 86
7.4 Rules of Differentiation 88
7.5 Applying Differentiation to Draw Graphs 90
7.6 Using Differential Calculus to Solve Problems 99
8 Linear Programming 108
8.1 Introduction 108
8.2 Terminology 108
8.3 Linear Programming and the Feasible Region 109
9 Geometry 117
9.1 Introduction 117
9.2 Circle Geometry 117
9.3 Coordinate Geometry 138
9.4 Transformations 143
10 Trigonometry 148
10.1 Introduction 148
10.2 Compound Angle Identities 148
10.3 Applications of Trigonometric Functions 155
10.4 Other Geometries 160
11 Statistics 164
11.1 Introduction 164
11.2 Normal Distribution 164
11.3 Extracting a Sample Population 166
11.4 Function Fitting and Regression Analysis 167
12 Combinations and Permutations 178
12.1 Introduction 178
12.2 Counting 178
12.3 Notation 179
12.4 Fundamental Counting Principle 179
12.5 Combinations 180
12.6 Permutations 182
12.7 Applications 184
Introduction to Book
The Language of Mathematics
The purpose of any language, like English or Zulu, is to make it possible for people to communicate.
All languages have an alphabet, which is a group of letters that are used to make up words. There are
also rules of grammar which explain how words are supposed to be used to build up sentences. This
is needed because when a sentence is written, the person reading the sentence understands exactly
what the writer is trying to explain. Punctuation marks (like a full stop or a comma) are used to further
clarify what is written.
Mathematics is a language, specifically it is the language of Science. Like any language, mathematics
has letters (known as numbers) that are used to make up words (known as expressions), and sentences
(known as equations). The punctuation marks of mathematics are the different signs and symbols that
are used, for example, the plus sign (+), the minus sign (— ), the multiplication sign (x), the equals sign
(=) and so on. There are also rules that explain how the numbers should be used together with the
signs to make up equations that express some meaning.
© See introductory video: VMinh at www.everythingmaths.co.za
Logarithms
2. 1 Introduction
In mathematics many ideas are related. We saw that addition and subtraction are related and that
multiplication and division are related. Similarly, exponentials and logarithms are related.
Logarithms, commonly referred to as logs, are the inverse of exponentials. The logarithm of a number
x in the base a is defined as the number n such that a n = x.
So, if a" = x, then:
log (x) = n
© See introductory video: VMgio at www.everythingmaths.co.za
(2.1)
2.2 Definition of Logarithms
The logarithm of a number is the value to which the base must be raised to give that number i.e. the
exponent. From the first example of the activity log 2 (4) means the power of 2 that will give 4. As
2 2 = 4, we see that
log 2 (4) = 2 (2.2)
The exponentialform is then 2 2 = 4 and the logarithmicform is log 2 4 = 2.
DEFINITION: Logarithms
If a" = x, then: log a (:r) = n, where a > 0; a ^ 1 and x > 0.
Activity.
Logarithm Symbols
Write the following out in words. The first one is done for you.
1 . log 2 (4) islogtothebase2of4
2 log 10 (14)
3 log 16 (4)
4. log, (8)
5. log,, (a;)
CHAPTER 2. LOGARITHMS
2.3
Activity:
Applying the definition
Find the value of:
1. log 7 343
2. log 2 8
3 log 4 i
4. log 10 1 000
Reasoning :
7 3 = 343
therefore, log 7 343 = 3
2.3 Logarithm Bases
Logarithms, like exponentials, also have a base and log 2 (2) is not the same as log 10 (2).
We generally use the "common" base, 10, or the natural base, e.
The number e is an irrational number between 2.71 and 2.72. It comes up surprisingly often in Math
ematics, but for now suffice it to say that it is one of the two common bases.
Extension:
Natural Logarithm
The natural logarithm (symbol In) is widely used in the sciences. The natural logarithm is to
the base e which is approximately 2.71828183 .... e, like tt and is an example of an irrational
number.
While the notation log 10 (a;) and log e (x) may be used, log 10 (x) is often written log(x) in Science and
log e (x) is normally written as ln(x) in both Science and Mathematics. So, if you see the log symbol
without a base, it means log 10 .
It is often necessary or convenient to convert a log from one base to another. An engineer might need
an approximate solution to a log in a base for which he does not have a table or calculator function,
or it may be algebraically convenient to have two logs in the same base.
Logarithms can be changed from one base to another, by using the change of base formula:
logtz
log„ x
log;, a
(2.3)
where b is any base you find convenient. Normally a and 6 are known, therefore log,, a is normally a
known, if irrational, number.
For example, change log 2 12 in base 10 is:
log, 12
logic 12
lo Sio 2
2.4
CHAPTER 2. LOGARITHMS
Activity:
Change of Base
Change the following to the indicated base:
1 . log 2 (4) to base 8
2. log 1(1 (14) to base 2
3. log 16 (4) to base 10
4. log x (8) to base y
5. log (x) to base x
© See video: VMgiq at www.everythingmaths.co.za
2.4 Laws of Logarithms
Just as for the exponents, logarithms have some laws which make working with them easier. These
laws are based on the exponential laws and are summarised first and then explained in detail.
log„(l) =
=
(2.4)
i°g Q 0) =
= 1
(2.5)
lo ga ( x ■ v) ~
= l°ga(z)+l°ga(2/)
(2.6)
*■(;) ■
= l°gaO) l°g„ (2/)
(2.7)
l°g„ (^J =
= &log a (a:)
(2.8)
log a (^) =
l0g o (!c)
b
(2.9)
2.5 Logarithm Law 1: log a 1 =
c
Since a
Then, log a (l)
For example,
and
1
by definition of logarithm in Equation 2.1
log 2 1 =
log,, 1 =
Activity:
Logarithm Law 7: log a 1 =
CHAPTER 2. LOGARITHMS
2.6
Simplify the following:
1. log 2 (l) + 5
2. log 10 (l) x 100
3. 3 x log 16 (l)
4. \og x (l) + 2xy
c l°Sv(l)
2.6 Logarithm Law 2: log a (a
For example,
and
r 1
Since a = a
Then, log a (a) = 1 by definition of logarithm in Equation 2.1
log, 2 = 1
log 25 25 = 1
Activity:
Logarithm Law 2: log„(a) = 1
Simplify the following:
1. log 2 (2) + 5
2. log 10 (10) x 100
3. 3 x log 16 (16)
4. logj,(a;) + 2xy
5.
l°g„(K)
Tip
When the base is 10 we do not need to state it. From the work done up to now, it is also useful to
summarise the following facts:
Useful to know and re
member
1 . log 1 =
2. log 10 = 1
3. log 100 = 2
4. log 1000 = 3
2.7
CHAPTER 2. LOGARITHMS
2.7 Logarithm Law 3:
log a (x2/) = log a (x) + log a (y)
The derivation of this law is a bit trickier than the first two. Firstly, we need to relate x and y to the
base a. So, assume that x = a m and y = a". Then from Equation 2.1, we have that:
log a (x) = m (2.10)
and logjy) = n (2.11)
This means that we can write:
log a (»•!/) = log a (a m .a")
= log a (a m+ ") (Exponential Law Equation (Grade 10))
= log a (a log " 0E)+log " (: " ) ) (From Equation 2.10 and Equation 2.1 1)
= log (x) + log a (y) (From Equation 2.1)
For example, show that log(10 . 100) = log 10 + log 100. Start with calculating the left hand side:
log(10.100) = log(1000)
= log(10 3 )
= 3
The right hand side:
log 10 + log 100 = 1 + 2
= 3
Both sides are equal. Therefore, log(10 . 100) = log 10 + log 100.
Activity.
Logarithm Law 3: log Q (x . y) = log (jc) + log (y)
Write as separate logs:
1. log 2 (8x4)
2. log 8 (10 x 10)
3 log l6 (xy)
4. \og z {2xy)
5 log, fe 2 )
2.8 Logarithm Law 4:
lQ ga (J) = lQ ga(^)  l0g a (2/)
EMCI
The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an exercise.
For example, show that log(^jjj) = log 10 — log 100. Start with calculating the left hand side:
1
logClO 1 )
1
logf^)
s \100J
ll «
CHAPTER 2. LOGARITHMS
2.9
The right hand side:
log 10  log 100 = 12
= 1
Both sides are equal. Therefore, log(j^) = log 10 — log 100.
Activity:
Logarithm Law 4: log a (  J = log a (a;)  log„(y)
Write as separate logs:
1. log 2 ()
2 log 8 (^)
3 log 16 (f)
4 log 2 (f)
5 log x ()
2.9 Logarithm Law 5:
lo ga(^) = bhg a {x
Once again, we need to relate x to the base a. So, we let x = a m . Then,
loga^') = l°ga (( a m ) b )
= log a (a m ' ) (Exponential Law in Equation (Grade 10))
But, m = log a (x) (Assumption that x = a m )
.; logjz 6 ) = ]og„(a' to « W )
= b.log a (x) (Definition of logarithm in Equation 2.1)
For example, we can show that log 2 (5 3 ) = 31og 2 (5).
log 2 (5 3 ) = log 2 (5.5.5)
log 2 5 + log 2 5 + log 2 5 (■.■ log a (:r . y) = log Q (a . a ))
3 log, 5
Therefore, log 2 (5 3 ) = 31og 2 (5).
Activity:
Logarithm Law 5: log a (z 6 ) = blog a (x)
Simplify the following:
1. log 2 (8 4 )
2. log 8 (10 10 )
3. log 16 (x")
4 iog z (y x )
5 log,(2/ 2x )
2.10
CHAPTER 2. LOGARITHMS
2.10 Logarithm Law 6:
The derivation of this law is identical to the derivation of Logarithm Law 5 and is left as an exercise.
For example, we can show that log 2 (v / 5) = ° s 3 .
log 2 (^5) = log 2 (5*)
1
3
log 2 5
log 2 5 (■.' log^a/") = 61og a (:r))
Therefore, log^v^) = l£ f 5 .
Activity:
Logarithm Law 6: log a {J/x)
logq(^)
Tip
The final answer doesn't
have to look simple.
Simplify the following:
1. log 2 (^8)
2. log 8 ( Vld)
3 log 16 (^/z)
4 log 2 (^)
® See video: VMgjI at www.everythingmaths.co.za
Example 1: Simplification of Logs
QUESTION
Simplify, without use of a
calculator:
3 log 3 + log 125
SOLUTION
Step 1 : Try to write any quantities as exponents
125 can be written as 5 3 .
Step 2 : Simplify
10
CHAPTER 2. LOGARITHMS 2.10
3 log 3 + log 125 = 3 log 3 + log 5 3
= 31og3 + 31og5 ■ : \og a {x b ) = b\og a {x)
= 3 log 15 (Logarithm Law 3)
Step 3 : Final Answer
We cannot simplify any further. The final answer is:
3 log 15
Example 2: Simplification of Logs
QUESTION
Simplify, without use of a calculator:
8^ +log 2 32
SOLUTION
Step 1 : Try to write any quantities as exponents
8 can be written as 2 3 . 32 can be written as 2 5 .
Step 2 : Rewrite the question using the exponential forms of the numbers
8* +log 2 32 = (2 3 )i +log 2 2 5
Step 3 : Determine which laws can be used.
We can use:
Step 4 : Apply log laws to simplify
(2 3 )I+log 2 2 5 = (2) 3x I+51og 2 2
Step 5 : Determine which laws can be used.
We can now use log a a = 1
Step 6 : Apply log laws to simplify
(2) 2 + 5 log 2 2 = 2 2 + 5(1) = 4 + 5 = 9
Step 7 : Final Answer
The final answer is:
83 +log 2 32 = 9
11
2.11
CHAPTER 2. LOGARITHMS
Example 3: Simplify to one log
QUESTION
Write 2 log 3 + log 2  log 5
as the logarithm
of a singl
e number.
SOLUTIOS
Step 7 :
Reverse law 5
2 log 3 + log 2
 log 5 =
log3 2
+ log2
log 5
Step 2 :
Apply laws 3 and 4
= log(3 2 x 2 = 5)
Step 3 :
Write the final
= log 3,6
answer
2. / / Solving Simple Log Equations
In Grade 10 you solved some exponential equations by trial and error, because you did not know the
great power of logarithms yet. Now it is much easier to solve these equations by using logarithms.
For example to solve x in 25 x = 50 correct to two decimal places you simply apply the following
reasoning. IftheLHS = RHS then the logarithm of the LHS must be equal to the logarithm of the RHS.
By applying Law 5, you will be able to use your calculator to solve for x.
Example 4: Solving Log equations
QUESTION
Solve for x: 25* = 50 correct to two decimal places.
12
CHAPTER 2. LOGARITHMS 2.11
SOLUTION
Step 1 : Taking the log of both sides
log 25 x = log 50
Step 2 : Use Law 5
x log 25 = log 50
Step 3 : Solve for x
x = log 50 — log 25
x = 1,21533...
Step 4 : Round off to required decimal place
x = 1,22
In general, the exponential equation should be simplified as much as possible. Then the aim is to
make the unknown quantity (i.e. x) the subject of the equation.
For example, the equation
is solved by moving all terms with the unknown to one side of the equation and taking all constants to
the other side of the equation
2 X . 2 2 = 1
1
2 = 2^
Then, take the logarithm of each side.
log (2*) = log(l
slog (2) = log(2 2 )
x log (2) = 2 log (2) Divide both sides by log (2)
.'. x = 2
2+2 = 2° = 1 /
Substituting into the original equation, yields
Similarly, 9 (1 ~ 2a,) = 3 4 is solved as follows:
g(l2x) _ g4
o2(l2i) n4
324* _ 3 4 take the  g ar j tnm f both sides
log(3 2  4 *) = log(3 4 )
(2  ix) log(3) = 41og(3) divide both sides by log(3)
2  Ax = 4
Ax = 2
1
:.x = 
Substituting into the original equation, yields
g(l2(^l)) _ g(l + l) _ 3 2(2) _ g4 j
13
2.11 CHAPTER 2. LOGARITHMS
Example 5
: Exponential Equation
QUESTION
Solve for x ir
7 , 5 (3*+3) =35
SOLUTION
Step 1 :
Identify the base with x as an exponent
There are two possible bases: 5 and 7. x is an exponent of 5.
Step 2 :
Eliminate the base with no x
In order to eliminate 7, divide both sides of the equation by 7 to
5 (3*+3) = 5
give:
Step 3 :
7a/re the logarithm of both sides
log(5 (te+3) ) = log(5)
Step 4 :
Apply the log laws to make x the subject of the equation.
(3x + 3) log(5) = log(5) divide both sides of the equation by
log(5)
3x + 3 = 1
Zx = 2
2
X = —3
Step 5 :
Substitute into the original equation to check answer.
7 .5«3x> + 3) =7 _ 5 (2 + 3) =7 5 1 =35 /
Exercise 21
Solve for x:
1 . log 3 x = 2
2. I0'°s 27 = x
3. 3 2x  1 = 2i lx  1
\fc\ More practice yr) video solutions Cf) or help at www.everythingmaths.co.za
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It
CHAPTER 2. LOGARITHMS
2.12
2.12 Logarithmic Applications in
the Real World
Logarithms are part of a number of formulae used in the Physical Sciences. There are formulae that
deal with earthquakes, with sound, and pHlevels to mention a few. To work out time periods is growth
or decay, logs are used to solve the particular equation.
Example 6: Using the growth formula
QUESTION
A city grows 5% every two years.
How long will it take for the city to triple its size?
SOLUTION
Step 1
Use the formula
A = P(l + i) n As
ume P = x, then A = 3x. For this example
n represents a
period of 2 years, therefore the n is halved for this purpose.
Step 2
Substitute information given into formula
3
= (1,05)5
log 3
= — x log 1,05 (using Law 5)
n
= 2 log 3 H log 1,05
n
= 45,034
Step 3
Final answer
So it will take approximately 45 years for the population to triple
in size.
Exercise 22
1. The population of a certain bacteria is expected to grow exponentially at a rate of 15% every
hour. If the initial population is 5 000, how long will it take for the population to reach 100 000?
2. Plus Bank is offering a savings account with an interest rate if 10% per annum compounded
monthly. You can afford to save R300 per month. How long will it take you to save R20 000?
(Give your answer in years and months)
2A2
CHAPTER 2. LOGARITHMS
A"j More practice Crj video solutions Cf) or help at www.everythingmaths.co.za
(l.)Olbr (2.)01bs
Example 7: Logs in Compound Interest
QUESTION
1 have R12 000 to invest. 1 need the money
to grow to at least R30 000. If it is invested at a
compound interest rate of 13% per annum,
for how long (in full years) does my investment
need to grow?
SOLUTION
Step 1 : The formula to use
A =
P(l + i) n
Step 2 : Substitute and solve for n
30000 <
12000(1 + 0,13)"
1,13" >
5
2
nlog(l,13) >
log(2,5)
n >
log(2,5)log(l,13)
n >
7,4972 . . .
Step 3 : Determine the final answer
In this case we round up, because 7 years will not yet deliver the required
R30 000. The investment need to stay in the bank for at least 8 years.
Chapter 2
End of Chapter Exercises
1 . Show that
lo Sa
;.W i°g„(y)
16
CHAPTER 2. LOGARITHMS 2.12
2. Show that
io Sa m = ^
3. Without using a calculator show that:
, 75 „. 5 . 32 , „
log 2 toe h log = log 2
6 16 6 9 & 243 &
4. Given that 5 n = x and n = log 2 y
(a) Write y in terms of n
(b) Express log 8 4y in terms of n
(c) Express 50" +1 in terms of x and y
5. Simplify, without the use of a calculator:
(a) 8 1 +log 2 32
(b) log 3 9  log 5 V5
1
^(ft4ft)' +1 ^ 92 ' 12
6. Simplify to a single number, without use of a calculator:
,„„ log 32 log 8
(a) log 5 125 + 2 — — S
log8
(b) log 3 log 0,3
7. Given: log 3 6 = a and log 6 5 = b
(a) Express log 3 2 in terms of a.
(b) Hence, or otherwise, find log 3 10 in terms of a and b.
8. Given: pq k = qp^ 1
Prove: k = 1 — 2 log g p
9. Evaluate without using a calculator: (log 7 49) 5 + log 5 I —  ) — 13 log 9 1
10. If log 5 = 0,7, determine, without using a calculator:
(a) log 2 5
(b) 10 1 ' 4
11. Given: M = log 2 (x + 3) + log 2 (x  3)
(a) Determine the values of x for which M is defined.
(b) Solve for x if M = 4.
/ \logx
12. Solve: ( x a ) = I0x 2 (Answer(s) may be left in surd form, if necessary.)
13. Find the value of (log 27 3) 3 without the use of a calculator.
14. Simplify By using a calculator: log 4 8 + 21og 3 \/27
15. Write log 4500 in terms of a and b if 2 = 10" and 9 = 10 6 .
5 2006 _ 5 2004 + 24
16. Calculate:
1
1 7. Solve the following equation for x without the use of a calculator and using the fact
that\/lOP» 3,16 :
21og(x + l)= „, 6 , t v 1
log(x + 1)
18. Solve the following equation for x: G bx = 66 (Give answer correct to two decimal
places.)
17
2.12 CHAPTER 2. LOGARITHMS
More practice Crj video solutions f 9) or help at www.everythingmaths.co.za
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(7.)01bz (8.)01c0 (9.)01c1 (10.)01c2 (1 1.) 01 c3 (12.)01c4
(13.)01c5 (14.) 01c6 (15.)01c7 (16.)01c8 (17.)01c9 (18.) Olca
is
Sequences and Series
3. 1 Introduction
In this chapter we extend the arithmetic and quadratic sequences studied in earlier grades, to geometric
sequences. We also look at series, which are the summing of the terms in sequences.
© See introductory video: VMgjm at www.everythingmaths.co.za
3.2 Arithmetic Sequences
The simplest type of numerical sequence is an arithmetic sequence.
DEFINITION: Arithmetic Sequence
An arithmetic (or linear) sequence is a sequence of numbers in which each new term
is calculated by adding a constant value to the previous term
For example,
1; 2; 3; 4; 5; 6; . . .
is an arithmetic sequence because you add 1 to the current term to get the next term:
first term
second term
third term
2 = 1 + 1
3 = 2+1
n th term: n = (n  1) + 1
Activity:
Common Difference
Find the constant value that is added to get the following sequences and write out the next
5 terms.
1. 2; 6; 10; 14; 18; 22;...
2. 5; 3;  1; 1; 3; . . .
3. 1; 4; 7; 10; 13; 16;...
4. 1; 10; 21; 32; 43; 54;...
19
3.2
CHAPTER 3. SEQUENCES AND SERIES
5. 3; 0; 3; 6; 9; 12;
General Equation for the n th Term of an
Arithmetic Sequence
EMCP
More formally, the number we start out with is called ai (the first term), and the difference between
each successive term is denoted d, called the common difference.
The general arithmetic sequence looks like:
ai
=
fli
a 2
=
Qi + d
03
=
0,2 + d = (qi + d) + d = a\ + 2d
cu
=
ai + d = (qi + 2d) + d = a\ + 3d
a n
=
ai + d . (n — 1)
Thus, the equation for the ri"'term is:
a„ = ai + d. (n — 1)
(3.1)
Given ai and the common difference, d, the entire set of numbers belonging to an arithmetic sequence
can be generated.
Tip
Test for Arithmetic Se
quences
DEFINITION: Arithmetic Sequence
An arithmetic (or linear) sequence is a sequence of numbers in which each new term
is calculated by adding a constant value to the previous term:
a„ = a„_i + d
(3.2)
where
a„ represents the new term, the n" l term, that is calculated;
a„_i represents the previous term, the (n — l)"*term;
d represents some constant.
A simple test for an arithmetic sequence is to check that the difference between consecutive terms is
constant:
ai — ai = a,3 — a 2 = a n — a n \ = d
(3.3)
This is quite an important equation, and is the definitive test for an arithmetic sequence. If this condi
tion does not hold, the sequence is not an arithmetic sequence.
2ll
CHAPTER 3. SEQUENCES AND SERIES
3.3
Extension:
Plotting a graph of terms in an arithmetic sequence
Plotting a graph of the terms of a sequence sometimes helps in determining the type of se
quence involved. For an arithmetic sequence, plotting o„ vs. n results in:
a n = Qi + d(n — 1)
(Note that the graph will only be a straight line if the sequence is arithmetic.)
3.3 Geometric Sequences
DEFINITION: Geometric Sequences
A geometric sequence is a sequence of numbers in which each new term (except for
the first term) is calculated by multiplying the previous term by a constant value.
This means that the ratio between consecutive numbers in the geometric sequence is a constant. We
will explain what we mean by ratio after looking at the following example.
21
3.3
CHAPTER 3. SEQUENCES AND SERIES
Example  A Flu Epidemic
EMCR
Extension:
What is influenza?
Influenza (commonly called "flu") is caused by the influenza virus, which infects the respiratory
tract (nose, throat, lungs). It can cause mild to severe illness that most of us get during winter
time. The main way that the influenza virus is spread is from person to person in respiratory
droplets of coughs and sneezes. (This is called "droplet spread".) This can happen when
droplets from a cough or sneeze of an infected person are propelled (generally, up to a metre)
through the air and deposited on the mouth or nose of people nearby. It is good practise to
cover your mouth when you cough or sneeze so as not to infect others around you when you
have the flu.
Assume that you have the flu virus, and you forgot to cover your mouth when two friends came to visit
while you were sick in bed. They leave, and the next day they also have the flu. Let's assume that they
in turn spread the virus to two of their friends by the same droplet spread the following day. Assuming
this pattern continues and each sick person infects 2 other friends, we can represent these events in
the following manner:
A A A. A.
ijjiijjiffijjiijjiijpijji
Figure 3.1 : Each person infects two more people with the flu virus.
Again we can tabulate the events and formulate an equation for the general case:
Day, n
Number of newlyinfected people
1
2=2
2
4 =2x2 = 2x2'
3
8 =2x4 = 2x2x2 = 2x2'
4
16 = 2x8 = 2x2x2x2 = 2x2^
5
32 = 2 x 16 = 2 x 2 x 2 x 2 x 2 = 2 x 2 4
n
= 2x2x2x2xx2 = 2x 2"~ x
The above table represents the number of newlyinfected people after n days since you first infected
your 2 friends.
22
CHAPTER 3. SEQUENCES AND SERIES
3.3
You sneeze and the virus is carried over to 2 people who start the chain (on = 2). The next day, each
one then infects 2 of their friends. Now 4 people are newlyinfected. Each of them infects 2 people the
third day, and 8 people are infected, and so on. These events can be written as a geometric sequence:
2; 4; 8; 16; 32; ...
Note the common ratio (2) between the events. Recall from the linear arithmetic sequence how the
common difference between terms was established. In the geometric sequence we can determine the
common ratio, r, from
a >
0.2
Or, more generally,
(3.4)
(3.5)
Activity:
Common Ratio of Geometric Sequence
Determine the common ratio for the following geometric sequences:
1. 5; 10; 20; 40; 80;...
i I I I
Z  2> 4' 8' ' "
3. 7; 28; 112; 448;...
4. 2; 6; 18; 54;...
5. 3; 30;  300; 3000; . . .
General Equation for the n th Term of a Ge
ometric Sequence
EMCS
From the flu example above we know that ai = 2 and r = 2, and we have seen from the table that the
n"*term is given by a„ = 2 x 2 n_1 . Thus, in general,
a„ = ai . r
(3.6)
where ai is the first term and r is called the common ratio.
So, if we want to know how many people are newlyinfected after 1 days, we need to work out aw:
nl
a„ = a\ .r
a w = 2 x 2 10  1
= 2 x 2 9
= 2 x 512
= 1024
That is, after 10 days, there are 1 024 newlyinfected people.
23
3.3
CHAPTER 3. SEQUENCES AND SERIES
Or, after how many days would 16 384 people be newlyinfected with the flu virus?
nl
a„
=
a\ . r
16 384
=
2 x 2' 1
384^2
=
2 „i
8 192
=
2 nl
2 13
=
2 „l
13
=
71 1
n
=
11
That is, after 14 days 16 384 people are newlyinfected.
Activity:
General Equation of Geometric Sequence
Determine the formula for the n"'term of the following geometric sequences:
1. 5; 10; 20; 40; 80;...
7 I I I
*■< 2 ' 4' 8' ' ' '
3. 7; 28; 112; 448; . . .
4. 2,6,18,54,...
5. 3; 30;  300; 3000; . . .
Exercise 31
1 . What is the important characteristic of an arithmetic sequence?
2. Write down how you would go about finding the formula for the n th term of an arithmetic
sequence?
3. A single square is made from 4 matchsticks. To make two squares in a row takes 7 matchsticks,
while three squares in a row takes 10 matchsticks. Determine:
(a) the first term
(b) the common difference
(c) the formula for the general term
(d) how many matchsticks are in a row of 25 squares
4. 5; x; y is an arithmetic sequence and x; y; 81 is a geometric sequence. All terms in the
sequences are integers. Calculate the values of x and y.
(A" 1 ) More practice (►) video solutions C 7} or help at www.everythingmaths.co.za
(1.)01cb (2.)01cc (3.)01cd (4.) 01ce
21
CHAPTER 3. SEQUENCES AND SERIES
3.4
3.4 Recursive Formulae for
Sequences
When discussing arithmetic and quadratic sequences, we noticed that the difference between two
consecutive terms in the sequence could be written in a general way.
For an arithmetic sequence, where a new term is calculated by taking the previous term and adding a
constant value, d:
a„ = a n i + d
The above equation is an example of a recursive equation since we can calculate the n'^term only by
considering the previous term in the sequence. Compare this with Equation (3.1),
a„ = oi + d. (n — 1)
(3.7)
where one can directly calculate the n"'term of an arithmetic sequence without knowing previous
terms.
For quadratic sequences, we noticed the difference between consecutive terms is given by (??):
a„  a„_i = D . (n — 2) + d
Therefore, we rewrite the equation as
o„ = Oni + D . (n  2) + d (3.8)
which is then a recursive equation for a quadratic sequence with common second difference, D.
Using (3.5), the recursive equation for a geometric sequence is:
a„ = r . a.ni (3.9)
Recursive equations are extremely powerful: you can work out every term in the series just by knowing
previous terms. As you can see from the examples above, working out a n using the previous term a„_i
can be a much simpler computation than working out a„ from scratch using a general formula. This
means that using a recursive formula when using a computer to work out a sequence would mean the
computer would finish its calculations significantly quicker.
Activity:
Recursive Formula
Write the first five terms of the following sequences, given their recursive formulae:
1 . a n = 2a n i + 3, cl\ = 1
2. a n = On— i,cn — 11
3. a n = 2a n _ 1 ,ai = 2
Extension:
The Fibonacci Sequence
Consider the following sequence:
0; 1; 1; 2; 3; 5; 8; 13; 21; 34;
(3.10)
The above sequence is called the Fibonacci sequence. Each new term is calculated by adding
3.5
CHAPTER 3. SEQUENCES AND SERIES
the previous two terms. Hence, we can write down the recursive equation:
a„ = a„i + a n 2 (3.11)
3.5 Series
In this section we simply work on the concept of adding up the numbers belonging to arithmetic and
geometric sequences. We call the sum of any sequence of numbers a series.
Some Basics
EMCV
If we add up the terms of a sequence, we obtain what is called a series. If we only sum a finite number
of terms, we get a fin ite series. We use the symbol S„ to mean the sum of the first n terms of a sequence
{ai; a2; 03; . . . ;a n }:
On
ai + 0.2 + a 3 + ■ ■ ■ + a„
(3.12)
For example, if we have the following sequence of numbers
1; 4; 9; 25; 36; 49; ...
and we wish to find the sum of the first four terms, then we write
Si = 1 + 4 + 9 + 25 = 39
The above is an example of a finite series since we are only summing four terms.
If we sum infinitely many terms of a sequence, we get an infinite series:
Son = ai + 02 + a 3 + . . .
(3.13)
Sigma Notation
EMCW
In this section we introduce a notation that will make our lives a little easier.
A sum may be written out using the summation symbol J^ • This symbol is sigma, which is the capital
letter "S" in the Greek alphabet. It indicates that you must sum the expression to the right of it:
Y.°
Om + Om+1 + ■ ■ ■ + Un1 + O n
(3.14)
where
is the index of the sum;
2(i
CHAPTER 3. SEQUENCES AND SERIES 3.5
• to is the lower bound (or start index), shown below the summation symbol;
• n is the upper bound (or end index), shown above the summation symbol;
• a; is a term of a sequence.
The index i increases from m to n in steps of 1.
If we are summing from i = 1 (which implies summing from the first term in a sequence), then we can
use either S„ or ^notation since they mean the same thing:
n
S„ = ^^ o,i = 0.1 + a2 + ■ ■ ■ + a n (3.15)
For example, in the following sum,
5
£*
we have to add together all the terms in the sequence a % = i from i = 1 up until i = 5:
5
^j= 1 + 2 + 3 + 4 + 5 = 15
i i
Examples
1.
£*
2 + 4 + 8 + 16 + 32 + 64
126
2.
J2(3x l ) = 3x 3 + 3x 4 + ■ ■ ■ + 3x g + 3x
for any value x.
Some Basic Rules for Sigma Notation
1 . Given two sequences, a, and 6,,
n n
Y^{a t + bi) = Y. a ' + 12 bi (3  16)
i=l i=l t=l
2. For any constant c that is not dependent on the index i,
n
> c . a, = c . ai + c . a2 + c . a,s + ■ ■ ■ + c . a n
,. i
= c (ai + a 2 + 03 + ■ ■ ■ + a„)
n
= c^a, (3.17)
27
3.6
CHAPTER 3. SEQUENCES AND SERIES
Exercise 32
1 . What is J2 2 ?
2. Determine J2 *•
3. Expand 2~Z *■
4. Calculate the value of a if:
^a.2 fc_1 = 28
f/VJ More practice CrJ video solutions f9 J or help at www.everythingmaths.co.za
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3.6 Finite Arithmetic Series
Remember that an arithmetic sequence is a sequence of numbers, such that the difference between
any term and the previous term is a constant number, d, called the constant difference:
a„ = ai + d (n — 1)
(3.18)
where
• n is the index of the sequence;
• a„ is the n^term of the sequence;
• ai is the first term;
• d is the common difference.
When we sum a finite number of terms in an arithmetic sequence, we get a finite arithmetic series.
A simple arithmetic sequence is when ai = 1 and d = in the general form (3.18); in other words all
the terms in the sequence are 1:
ai = ai + d (i — 1)
= l + 0.(il)
= 1
{a,} = {1; 1; 1; 1; 1; ...}
If we wish to sum this sequence from i = 1 to any positive integer n, we would write
n n
^ ai = ^l = l + l + H hi (n times)
2S
CHAPTER 3. SEQUENCES AND SERIES
3.6
Because all the terms are equal to 1, it means that if we sum to n we will be adding ?inumber of l's
together, which is simply equal to n:
Ei
(3.19)
Another simple arithmetic sequence is when <n = 1 and d = I, which is the sequence of positive
integers:
a; = ai + d (i — 1)
= 1 + 1 . (*  1)
= i
{a,} = {1; 2; 3; 4; 5; ...}
If we wish to sum this sequence from i = 1 to any positive integer re, we would write
Y^i= 1 + 2 + 3
(3.20)
This is an equation with a very important solution as it gives the answer to the sum of positive integers.
We first write S„ as a sum of terms in ascending order:
S„ = 1 + 2 + h (n  1) + n
We then write the same sum but with the terms in descending order:
S n = n + {n  1) + h 2 + 1
(3.21)
(3.22)
We then add corresponding pairs of terms from Equations (3.21) and (3.22), and we find that the sum
for each pair is the same, (n + 1):
2 S n = (n + 1) + (n + 1) + • • • + (n + 1) + (n + 1) (3.23)
We then have renumber of (n + l)terms, and by simplifying we arrive at the final result:
2 S n = re (re + 1)
Sn = (n + l)
S n = ^2 i =  (n + 1)
(3.24)
Note that this is an example of a quadratic sequence.
FACT
Mathematician, Karl
Friedrich Gauss, discov
ered this proof when
he was only 8 years
old. His teacher had
decided to give his
class a problem which
would distract them
for the entire day by
asking them to add
all the numbers from
1 to 100. Young Karl
realised how to do this
almost instantaneously
and shocked the teacher
with the correct answer,
5050.
If we wish to sum any arithmetic sequence, there is no need to work it out termforterm. We will now
determine the general formula to evaluate a finite arithmetic series. We start with the general formula
20
3.6
CHAPTER 3. SEQUENCES AND SERIES
for an arithmetic sequence and sum it from i = 1 to any positive integer n:
n n
Y^a t = ^[oi+d(il)]
i=l i=l
n
= / (ai + di — d)
n
= Yl K ai  d )+ d{ \
n n
= $^(aid) + X)(di)
i=l i=l
n n
= J2 ( ai  d ) + d J2 l
i a tin , , .
= (ai — d)n+ — (n + 1)
=  (2oi  2d + dn + d)
= — (2oi + dn — d)
= ^[2oi+d(nl)]
So, the general formula for determining an arithmetic series is given by
Sn = J2 [ai+d{il)] = ^ [2oi+d(nl)]
(3.25)
For example, if we wish to know the series S20 for the arithmetic sequence a; = 3 + 7 (i — 1), we
could either calculate each term individually and sum them:
^'20
£[3 + 7(il)
= 3 + 10 + 17 + 24 + 31 + 38 + 45 + 52 +
59 + 66 + 73 + 80 + 87 + 94 + 101 +
108+ 115 + 122+ 129 + 136
= 1390
or, more sensibly, we could use Equation (3.25) noting that 01 = 3, d = 7 and n = 20 so that
20
S 20 = X> + 7(il)]
= f [2.3 + 7(201)]
= 1390
This example demonstrates how useful Equation (3.25) is.
Exercise 33
1 . The sum to n terms of an arithmetic series is S n = — (In + 15).
(a) How many terms of the series must be added to give a sum of 425?
(b) Determine the (5 th term of the series.
2. The sum of an arithmetic series is 100 times its first term, while the last term is 9 times the first
term. Calculate the number of terms in the series if the first term is not equal to zero.
30
CHAPTER 3. SEQUENCES AND SERIES
3.7
3. The common difference of an arithmetic series is 3. Calculate the value of n for which the n th
term of the series is 93, and the sum of the first n terms is 975.
4. The sum of n terms of an arithmetic series is 5n 2 — lln for all values of n. Determine the
common difference.
5. The third term of an arithmetic sequence is —7 and the 7 th term is 9. Determine the sum of the
first 51 terms of the sequence.
6. Calculate the sum of the arithmetic series 4 + 7 + 10 + ■ ■ ■ + 901.
A" 1 ) More practice CwJ video solutions Qfj or help at www.everythingmaths.co.za
(1 .) Olcj (2.)01ck (3.) 01cm (4.) 01 en (5.) 01cp (6.) Olcq
3.7 Finite Squared Series
When we sum a finite number of terms in a quadratic sequence, we get a finite quadratic series. The
general form of a quadratic series is quite complicated, so we will only look at the simple case when
D = 2 and d = (02 — ai) = 3, where D is the common second difference and d is the first difference.
This is the sequence of squares of the integers:
■ 2
oi = 1
{a,} = {l 2 ; 2 2 ; 3 2 ; 4 2 ; 5 2 ; 6 2 ; . . .}
= {1; 4; 9; 16; 25; 36; ...}
If we wish to sum this sequence and create a series, we write
n
Sn = J2 * 2 = 1 + 4 + 9 + ■■■ + n
1 1
which can be written, in general, as
S„ = ^Y = ^(2n + l)(n + l)
1=1
The proof for Equation (3.26) can be found under the Extension block that follows:
(3.26)
Extension:
Derivation of the Finite Squared Series
We will now prove the formula for the finite squared series:
S„ = ^y = l + 4 + 9 +  + n 2
We start off with the expansion of (k + 1)'
(fc + 1) 3
(jfc+l) 3 fc 3 = 3/c 2 + 3fc + l
31
3.8
CHAPTER 3. SEQUENCES AND SERIES
k= 1
: 2 3
 l 3 = 3(1) 2 +3(1) + 1
k = 2 : 3 3
2 3 = 3(2) 2 +3(2) + l
k = 3 : 4 3
3 3 = 3(3) 2 +3(3) + l
k = n : (n
+ 1) 3 n 3 = 3n 2 + 3n + l
If we add all the terms on the right and left, we arrive at
(„ + l) 3  1 =
n
J2 ( 3 * 2 + 3i + 1)
n" + 3n~ + 'in + 1 — 1 =
71 n n
sE^E^ + E 1
i=l i=l i=l
n + 3n + 3n =
n
3^* 2 + T (n+l)+n
1 = 1
n
E* 2 =
I 1
i [n 3 + 3n 2 + 3n  ^ (n + 1)  n]
=
1 /• 3 i q 2 , 3 2 3 .
— (n + 3n + 3?i — — n — — n — n)
=
l f J,3 2,1 s
— in H — n H — n.)
3 V 2 2 '
=
 (2n + 3n + 1)
6
Therefore,
71
E* 2 =
£(2n + l)(n + l)
3.8 F/n/te Geometric Series
When we sum a known number of terms in a geometric sequence, we get a finite geometric series.
We can write out each term of a geometric sequence in the general form:
a„ = ai . r
(3.27)
where
n is the index of the sequence;
a n is the n' h term of the sequence;
ai is the first term;
r is the common ratio (the ratio of any term to the previous term).
32
CHAPTER 3. SEQUENCES AND SERIES 3.8
By simply adding together the first n terms, we are actually writing out the series
S n = ai + ai r + ai r + ■ ■ ■ + ai r n ~~ + ai r n ~ (3.28)
We may multiply the above equation by r on both sides, giving us
rSn = air + air + air + ■ ■ ■ + ai r n + Qi r" (3.29)
You may notice that all the terms on the right side of (3.28) and (3.29) are the same, except the first
and last terms. If we subtract (3.28) from (3.29), we are left with just
rS„ — S„ = ai r n — ai
S n {rl) = ai(r n l)
Dividing by (r — 1) on both sides, we arrive at the general form of a geometric series:
•A _! ai (r n  1)
i = > ai.r = (3.30)
i. — ( r — 1
i=l
See video: VMgjq at www.everythingmaths.co.za
Exercise 34
1 . Prove that
L ^2^ , n1 a(lr")
a + ar + ar + ■ ■ • + ar = — —
(1r)
2. Given the geometric sequence 1; —3; 9; . . . determine:
(a) The 8 th term of the sequence
(b) The sum of the first eight terms of the sequence.
3. Determine:
4
,, i
4. Find the sum of the first 11 terms of the geometric series 6 + 3+ +  + ...
5. Show that the sum of the first n terms of the geometric series
54+18 + 6 +  + 5(i) n ~ 1
is given by 813 4 "".
6. The eighth term of a geometric sequence is 640. The third term is 20. Find the sum of the first 7
terms.
7. Solve for n; E8()' = 15§.
4=1
8. The ratio between the sum of the first three terms of a geometric series and the sum of the 4" 1 ,
5"' and 6"'terms of the same series is 8 : 27. Determine the common ratio and the first 2 terms
if the third term is 8.
f/Vj More practice f ►) video solutions C"?) or help at www.everythingmaths.co.za
(1.)01cr (2.) 01cs (3.)01ct (4.) 01cu (5.) 01cv (6.) 01cw
(7.) 01 ex (8.)01cy
33
3.9
CHAPTER 3. SEQUENCES AND SERIES
3.9 Infinite Series
Thus far we have been working only with finite sums, meaning that whenever we determined the
sum of a series, we only considered the sum of the first n terms. In this section, we consider what
happens when we add infinitely many terms together. You might think that this is a silly question 
surely the answer will be oo when one sums infinitely many numbers, no matter how small they are?
The surprising answer is that while in some cases one will reach oo (like when you try to add all the
positive integers together), there are some cases in which one will get a finite answer. If you don't
believe this, try doing the following sum, a geometric series, on your calculator or computer:
You might think that if you keep adding more and more terms you will eventually get larger and larger
numbers, but in fact you won't even get past 1  try it and see for yourself!
We denote the sum of an infinite number of terms of a sequence by
Y. a >
When we sum the terms of a series, and the answer we get after each summation gets closer and closer
to some number, we say that the series converges. If a series does not converge, then we say that it
diverges.
Infinite Geometric Series
EMCAC
There is a simple test for knowing instantly which geometric series converges and which diverges.
When r, the common ratio, is strictly between —1 and 1, i.e. — 1 < r < 1, the infinite series will
converge, otherwise it will diverge. There is also a formula for working out the value to which the
series converges.
Let's start off with Formula (3.30) for the finite geometric series:
I>i
,.«_! = oi (r"  1)
r1
Now we will investigate the behaviour of r" for — 1 < r < 1 as n becomes larger.
Take r = = :
n= 1
n = 2
n = 3
y2) 2
(i) 2 = 2
1
2
4^2
(k) 3 = k
1
2
1 = i <
2 8 ^
Since r is in the range — 1 < r < 1, we see that r" gets closer to as n gets larger.
.31
CHAPTER 3. SEQUENCES AND SERIES
3.9
Therefore,
On
'3 :^c
ai(r"
1)
r —
1
oi(0
1)
r —
1
— Ol
r 1
ai
for — 1 < r < 1
1 r
The sum of an infinite geometric series is given by the formula
for
1 < r < 1
where ax is the first term of the series and r is the common ratio.
See video: VMgly at www.everythingmaths.co.za
Exercise 35
(3.31)
1 . What does (§)" approach as n tends towards oo?
2. Given the geometric series:
2.(5) 5 +2.(5) 4 + 2.(5) ;i + ...
(a) Show that the series converges
(b) Calculate the sum to infinity of the series
(c) Calculate the sum of the first eight terms of the series, correct to two decimal places.
(d) Determine
oo
J2 2 ■ 5 6 ~"
correct to two decimal places using previously calculated results.
3. Find the sum to infinity of the geometric series 3+l +  +  + ...
4. Determine for which values of x, the geometric series
2 + f (x + 1) +  (x + if + . . .
will converge.
5. The sum to infinity of a geometric series with positive terms is 4 and the sum of the first two
terms is 2. Find a, the first term, and r, the common ratio between consecutive terms.
A" 1 ) More practice CrJ video solutions f?J or help at www.everythingmaths.co.za
(1.)01cz (2.) 01dO (3.) 01d1 (4.) 01d2 (5.) 01d3
Chapter 3
End of Chapter Exercises
;)5
3.9 CHAPTER 3. SEQUENCES AND SERIES
1 . lsl + 2 + 3 + 4 +  an example of a finite series or an infinite series?
2. Calculate
£3(i)*
3. If x + 1; x — 1; 2x — 5 are the first three terms of a convergent geometric series,
calculate the:
(a) Value of x.
(b) Sum to infinity of the series.
4. Write the sum of the first twenty terms of the series 6 + 3+ § + § + ••• in ^notation.
5. For the geometric series,
54+18 + 6 +  + 5() n  1
calculate the smallest value of n for which the sum of the first n terms is greater than
80.99.
CO
6. Determine the value of J2 12 (k) kl 
k = l
7. A new soccer competition requires each of 8 teams to play every other team once.
(a) Calculate the total number of matches to be played in the competition.
(b) If each of n teams played each other once, determine a formula for the total
number of matches in terms of n.
8. The midpoints of the opposite sides of square of length 4 units are joined to form 4
new smaller squares. This midpoints of the new smaller squares are then joined in
the same way to make even smaller squares. This process is repeated indefinitely.
Calculate the sum of the areas of all the squares so formed.
9. Thembi worked parttime to buy a Mathematics book which cost R29,50. On 1
February she saved Rl,60, and everyday saves 30 cents more than she saved the
previous day. (So, on the second day, she saved R1.90, and so on.) After how many
days did she have enough money to buy the book?
1 0. Consider the geometric series:
5 + 2i + l + ...
(a) If A is the sum to infinity and B is the sum of the first n terms, write down the
value of:
i. A
ii. B in terms of n.
(b) For which values of n is (A — B) < ^?
1 1 . A certain plant reaches a height of 118 mm after one year under ideal conditions in a
greenhouse. During the next year, the height increases by 12 mm. In each successive
year, the height increases by § of the previous year's growth. Show that the plant will
never reach a height of more than 150 mm.
n
12. Calculate the value of n if J2 (20  4a) = 20.
a=l
13. Michael saved R400 during the first month of his working life. In each subsequent
month, he saved 10% more than what he had saved in the previous month.
(a) How much did he save in the 7 th working month?
(b) How much did he save all together in his first 12 working months?
(c) In which month of his working life did he save more than Rl 500 for the first
time?
14. A man was injured in an accident at work. He receives a disability grant of R4 800 in
the first year. This grant increases with a fixed amount each year.
36
CHAPTER 3. SEQUENCES AND SERIES 3.9
(a) What is the annual increase if, over 20 years, he would have received a total of
R143 500?
(b) His initial annual expenditure is R2 600 and increases at a rate of R400 per year.
After how many years does his expenses exceed his income?
15. The Cape Town High School wants to build a school hall and is busy with fundraising.
Mr. Manuel, an exlearner of the school and a successful politician, offers to donate
money to the school. Having enjoyed mathematics at school, he decides to donate
an amount of money on the following basis. He sets a mathematical quiz with 20
questions. For the correct answer to the first question (any learner may answer), the
school will receive 1 cent, for a correct answer to the second question, the school
will receive 2 cents, and so on. The donations 1; 2; 4: . . . form a geometric sequence.
Calculate (Give your answer to the nearest Rand)
(a) The amount of money that the school will receive for the correct answer to the
20 th question.
(b) The total amount of money that the school will receive if all 20 questions are
answered correctly.
16. The common difference of an arithmetic series is 3. Calculate the values of n for
which the n th term of the series is 93 and the sum of the first n terms is 975.
1 7. The first term of a geometric sequence is 9, and the ratio of the sum of the first eight
terms to the sum of the first four terms is 97 : 81. Find the first three terms of the
sequence, if it is given that all the terms are positive.
18. (fe — 4); (fc + 1); m; 5fc is a set of numbers, the first three of which form an arithmetic
sequence, and the last three a geometric sequence. Find k and m if both are positive.
19. Given: The sequence 6 + p; 10 + p: 15 + p is geometric.
(a) Determine p.
(b) Show that the common ratio is f .
(c) Determine the 10"* term of this sequence correct to one decimal place.
20. The second and fourth terms of a convergent geometric series are 36 and 16, respec
tively. Find the sum to infinity of this series, if all its terms are positive.
n , c . , 5 k(k + l)
21. Eva uate: V —
fc = 2 z
22. S„ = An 1 + 1 represents the sum of the first n terms of a particular series. Find the
second term.
oo 12
23. Find p if: £ 27p fc = £ (243*)
fc=i t=i
24. Find the integer that is the closest approximation to:
102002 __ 1Q2002
25. In each case (substituting the values of x below), determine if the series J2( x + 2) p
P =i
converges. If it does, work out what it converges to:
(a)x = 5
(b) x = 5
26. Calculate: £ 5.4^
27. The sum of the first p terms of a sequence is p (p + 1). Find the 10 th term.
28. The powers of 2 are removed from the set of positive integers
1; 2; 3; 4; 5; 6; ... ; 1998; 1999; 2000
Find the sum of remaining integers.
37
3.9
CHAPTER 3. SEQUENCES AND SERIES
29. Observe the pattern below:
(a) If the pattern continues, find the number of letters in the column containing M's.
(b) If the total number of letters in the pattern is 361, which letter will be in the last
column.
30. The following question was asked in a test:
Find the value of 2 2005 + 2 2005 .
Here are some of the students' answers:
31
(a) Megan said the answer is 4 2005 .
(b) Stefan wrote down 2 4010 .
(c) Nina thinks itis2 2006 .
(d) Annatte gave the answer 2 2005x2005 .
Who is correct? ("None of them" is also a possibility.)
A shrub of height 110 cm is planted. At the end of the first year, the shrub is 120 cm
tall. Thereafter, the growth of the shrub each year is half of its growth in the previous
year. Show that the height of the shrub will never exceed 130 cm.
More practice (►) video solutions ({J or help at www.everythingmaths.co.za
(1.) 01d4
(7.) 01 da
(13.)01dg
(19.) 01dn
(25.) 01du
(31.) 01e0
(2.) 01 d5
(8.) 01 db
(14.) 01dh
(20.) 01 dp
(26.) 01 dv
(3.) 01d6
(9.) 01 dc
(15.) 01di
(21.) 01dq
(27.) 01 dw
(4.) 01d7
(10.) 01dd
(16.) 01dj
(22.) 01dr
(28.) 01 dx
(5.) 01d8
(11.) 01de
(17.) 01dk
(23.) 01ds
(29.) 01 dy
(6.) 01 d9
(12.) 01df
(18.) 01dm
(24.) 01dt
(30.) 01dz
:is
Finance
4. 1 Introduction
In earlier grades simple interest and compound interest were studied, together with the concept of
depreciation. Nominal and effective interest rates were also described. Since this chapter expands on
earlier work, it would be best if you revised the work done in Grades 10 and 1 1,
When you master the techniques in this chapter, you will be able to assess critically how to invest
your money when you start working and earning. And when you are looking at applying for a bond
from a bank to buy a home, you will confidently be able to get out the calculator and work out how
much you could actually save by making additional repayments. This chapter will provide you with
the fundamental concepts you will need to manage your finances.
© See introductory video: VMgmf at www.everythingmaths.co.za
4.2 Finding the Length of the
Investment or Loan
In Grade 1 1, we used the Compound Interest formula A = P(l + i) n to determine the term of the
investment or loan, by trial and error. Remember that P is the initial amount, A is the current amount,
i is the interest rate and n is the number of time units (number of months or years). So if we invest an
amount and know what the interest rate is, then we can work out how long it will take for the money
to grow to the required amount.
Now that you have learnt about logarithms, you are ready to work out the proper algebraic solution. If
you need to remind yourself how logarithms work, go to Chapter 2 (on Page 4).
The basic finance equation is:
A = P.(l + i) n
If you don't know what A, P, i and n represent, then you should definitely revise the work from Grade
10 and 11.
Solving for n:
A = P{l + i) n
(l + i)» = (A/P)
log((l+*) n ) = log(A/P)
nlog(l + i) = log(A/P)
n = log(A/P)/log(l + i)
Remember, you do not have to memorise this formula. It is very easy to derive any time you need
it. It is simply a matter of writing down what you have, deciding what you need, and solving for that
variable.
39
4.3
CHAPTER 4. FINANCE
Example 1: Term of Investment  Logarithms
QUESTION
Suppose we invested R3 500 into a savings account which pays 7,5% compound interest. After
an unknown period of time our account is worth R4 044,69. For how long did we invest the
money? How does this compare with the trial and error answer from Chapter ??.
SOLUTION
Step 7 :
Determine what is given and what is required
• P = R3 500
• i = 7,5%
. A = R4 044,69
We are required to find n
Step 2 :
Determine how to approach the problem
We know that:
A =
P(l + i) n
(1 + 0" =
(A/P)
bg((l + i)") =
log(A/P)
nlog(l + i) =
log(A/P)
n =
log(A/P)/log(l + i)
Step 3 :
Solve the problem
n = log(A/P)/log(l + »)
logf 4044 ' 69 ) 7 5
= , , 3 ™ ' Remember that: 7.5% ='„„= 0.075
log(l + 0.075) 100
= 2.0
Step 4 :
Write final answer
The R3 500 was invested for 2
years.
4.3 Series of Payments
By this stage, you know how to do calculations such as 'If I want Rl 000 in three years' time, how
much do I need to invest now at 10%?'
What if we extend this as follows: "If I want to draw Rl 000 next year, Rl 000 the next year and Rl 000
10
CHAPTER 4. FINANCE 4.3
after three years, how much do I need to initially put into a bank account earning 10% p.a. to be able
to afford to be able to do this?"
The obvious way of working that out is to work out how much you need now to afford the pay
ments individually and sum them. We'll work out how much is needed now to afford the payment of
Rl 000 in a year (= Rl 000 x (1,10) _1 =R909,09), the amount needed now for the following year's
Rl 000(=R1 000 x (1,10)~ 2 = R826,45) and the amount needed now for the Rl 000 after three years
(=R1 000 x (1,10)~ 3 =R751,31). Adding these together gives you the amount needed to afford all
three payments and you get R2 486,85.
So, if you put R2 486,85 into a 10% bank account now, you will be able to draw out Rl 000 in a year,
Rl 000 a year after that, and Rl 000 a year after that  and your bank account balance will decrease to
R0. You would have had exactly the right amount of money to do that (obviously!).
You can check this as follows:
Amount at Time (i.e. Now) = R2 486,85
Amount at Time 1 (i.e. a year later) = R2 486,85(1 + 10%) = R2 735,54
Amount after withdrawing Rl 000 = R2 735,54  Rl 000 = Rl 735,54
Amount at Time 2 (i.e. a year later) = Rl 735,54(1 + 10%) = Rl 909,09
Amount after withdrawing Rl 000 = Rl 909,09  Rl 000 = R909,09
Amount at Time 3 (i.e. a year later) = R909,09(l + 10%) = Rl 000
Amount after withdrawing Rl 000 = Rl 000  Rl 000 = R0
Perfect! Of course, for only three years, that was not too bad. But what if I asked you how much you
needed to put into a bank account now, to be able to afford R100 a month for the next 1 5 years. If you
used the above approach you would still get the right answer, but it would take you weeks!
There is  I'm sure you guessed  an easier way! This section will focus on describing how to work
with:
• annuities  a fixed sum payable each year or each month, either to provide a predetermined
sum at the end of a number of years or months (referred to as a future value annuity) or a fixed
amount paid each year or each month to repay (amortise) a loan (referred to as a present value
annuity).
• bond repayments  a fixed sum payable at regular intervals to pay off a loan. This is an example
of a present value annuity.
• sinking funds  an accounting term for cash set aside for a particular purpose and invested so that
the correct amount of money will be available when it is needed. This is an example of a future
value annuity.
Sequences and Series W emcag
Before we progress, you need to go back and read Chapter 3 (from Page 19) to revise sequences and
series.
In summary, if you have a series of n terms in total which looks like this:
a + ar + ar + ■ ■ ■ + ar = a[l + r + r + ■ ■ ■ r
this can be simplified as:
a{r n  1) , , ,
— useful when r > 1
r — 1
all — r")
— useful when < r < 1
1 — r
11
4.3 CHAPTER 4. FINANCE
Present Values of a Series of Payments W emcah
So having reviewed the mathematics of sequences and series, you might be wondering how this is
meant to have any practical purpose! Given that we are in the finance section, you would be right to
guess that there must be some financial use to all this. Here is an example which happens in many
people's lives  so you know you are learning something practical.
Let us say you would like to buy a property for R300 000, so you go to the bank to apply for a mortgage
bond. The bank wants it to be repaid by annually payments for the next 20 years, starting at end of
this year. They will charge you 15% interest per annum. At the end of the 20 years the bank would
have received back the total amount you borrowed together with all the interest they have earned from
lending you the money. You would obviously want to work out what the annual repayment is going to
be!
Let X be the annual repayment, i is the interest rate, and M is the amount of the mortgage bond you
will be taking out.
Time lines are particularly useful tools for visualising the series of payments for calculations, and we
can represent these payments on a time line as:
XX XXX
t t i ■ ' iii T c : h e Flows
12 18 19 20
Figure 4.1: Time line for an annuity (in arrears) of X for n periods.
The present value of all the payments (which includes interest) must equate to the (present) value of
the mortgage loan amount.
Mathematically, you can write this as:
M = X(l + i) 1 + X(l + €) 2 + X{1 + i) 3 + ■ ■ ■ + X(l + i) 20
The painful way of solving this problem would be to do the calculation for each of the terms above 
which is 20 different calculations. Not only would you probably get bored along the way, but you are
also likely to make a mistake.
Naturally, there is a simpler way of doing this! You can rewrite the above equation as follows:
M = Xlv 1 + v 2 + v 3 + h v 20 }
where v = (1 + «) _1 = 1/(1 + i)
Of course, you do not have to use the method of substitution to solve this. We just find this a useful
method because you can get rid of the negative exponents  which can be quite confusing! As an
exercise  to show you are a real financial whizz  try to solve this without substitution. It is actually
quite easy.
Now, the item in square brackets is the sum of a geometric sequence, as discussion in section 3. This
12
CHAPTER 4. FINANCE
4.3
can be rewritten as follows, using what we know from Chapter 3 of this text book:
v(l + v + v + • ■ ■ + v n ~ )
' \v n S
1 ~
lv n
l/v 1
l(i + Q
i
Note that we took out a common factor of v before using the formula for the geometric sequence.
So we can write:
M = X
(l(l + i)»)
This can be rewritten:
X
M
Mi
(i(i + Q")l
(l + i) n
So, this formula is useful if you know the amount of the mortgage bond you need and want to work
out the repayment, or if you know how big a repayment you can afford and want to see what property
you can buy.
For example, if I want to buy a house for R300 000 over 20 years, and the bank is going to charge me
15% per annum on the outstanding balance, then the annual repayment is:
.Y
Mi
1 (1 + i)""
R300 000x0,15
1(1 + 0,15) 20
R47 928.44
This means, each year for the next 20 years, I need to pay the bank R47 928,44 per year before I have
paid off the mortgage bond.
On the other hand, if I know I will have only R30 000 per year to repay my bond, then how big a house
can I buy? That is easy.
M
X
(i(i + 0)
R30 000
(1  (1,15) 20 )
0,15
R187 779,94
So, for R30 000 a year for 20 years, I can afford to buy a house of R187 800 (rounded to the nearest
hundred).
The bad news is that R187 800 does not come close to the R300 000 you wanted to pay! The good
news is that you do not have to memorise this formula. In fact , when you answer questions like this in
an exam, you will be expected to start from the beginning  writing out the opening equation in full,
showing that it is the sum of a geometric sequence, deriving the answer, and then coming up with the
correct numerical answer.
1.3
4.3 CHAPTER 4. FINANCE
Example 2: Monthly mortgage repayments
QUESTION
Sam is looking to buy his first flat, and has R15 000 in cash savings which he will use as a
deposit. He has viewed a flat which is on the market for R250 000, and he would like to work
out how much the monthly repayments would be. He will be taking out a 30 year mortgage
with monthly repayments. The annual interest rate is 11%.
SOLUTION
Step 1 : Determine what is given and what is needed
• Deposit amount = R15 000
The following is given: • Price of flat = R250 000
• interest rate, i = 11%
We are required to find the monthly repayment for a 30year mortgage.
Step 2 : Determine how to approach the problem
We know that:
X: M
(l(l + i)") l
In order to use this equation, we need to calculate M, the amount of the mortgage
bond, which is the purchase price of property less the deposit which Sam pays
upfront.
M = R250 000  R15 000
= R235 000
Now because we are considering monthly repayments, but we have been
given an annual interest rate, we need to convert this to a monthly interest rate,
ji2. (If you are not clear on this, go back and revise Section ??.)
(i + ii 2 r
= (i + »)
(1 + J12) 12
= 1,11
ii2
= 0,873459%
We know that the mortgage bond is for 30 years, which equates to 360
months.
Step 3 : Solve the problem
Now it is easy, we can just plug the numbers in the formula, but do not forget
that you can always deduce the formula from first principles as well!
X
M
r (l(l + i)") i
R235 000
r (l(1.00876459) 360 ) i
I 0,008734594 J
R2 146,39
Step 4 : Write the final answer
li
CHAPTER 4. FINANCE 4.3
That means that to buy a flat for R250 000, after Sam pays a R15 000 deposit,
he will make repayments to the bank each month for the next 30 years equal to
R2 146,39.
Example 3: Monthly mortgage repayments
QUESTION
You are considering purchasing a flat for R200 000 and the bank's mortgage rate is currently
9% per annum payable monthly. You have savings of R10 000 which you intend to use for
a deposit. How much would your monthly mortgage payment be if you were considering a
mortgage over 20 years.
SOLUTION
Step 1 : Determine what is given and what is required
The following is given:
• Deposit amount = R10 000
• Price of flat = R200 000
• Interest rate, i = 9%
We are required to find the monthly repayment for a 20year mortgage.
Step 2 : Determine how to approach the problem
We are considering monthly mortgage repayments, so it makes sense to use
months as our time period.
The interest rate was quoted as 9% per annum payable monthly, which
means that the monthly effective rate = 5J = 0,75% per month. Once we
have converted 20 years into 240 months, we are ready to do the calculations!
First we need to calculate M, the amount of the mortgage bond, which is
the purchase price of property minus the deposit which Sam pays upfront.
M = R200 000  R10 000
= R190 000
The present value of our mortgage payments X (which includes interest),
must equate to the present mortgage amount
M = X x (H0,75%) 1 +
X x (l + 0,75%r 2 +
X x (l + 0,75%r 3 +
X x (l + 0,75%r 4 + ...
X x (1 + 0,75%r 239 + X x (1 + 0,75%r 240
But it is clearly much easier to use our formula than work out 240 factors
and add them all up!
Step 3 : Solve the problem
15
4.3 CHAPTER 4. FINANCE
X x 111,14495 = R190 000
X = Rl 709,48
Step 4 : Write f/ie ffna/ answer
So to repay a R190 000 mortgage over 20 years, at 9% interest payable monthly,
will cost you Rl 709,48 per month for 240 months.
Show Me the Money
Now that you've done the calculations for the worked example and know what the monthly repayments
are, you can work out some surprising figures. For example, Rl 709,48 per month for 240 months
makes for a total of R410 275,20 (=R1 709,48 x 240). That is more than double the amount that you
borrowed! This seems like a lot. However, now that you've studied the effects of time (and interest) on
money, you should know that this amount is somewhat meaningless. The value of money is dependant
on its timing.
Nonetheless, you might not be particularly happy to sit back for 20 years making your Rl 709,48
mortgage payment every month knowing that half the money you are paying are going toward interest.
But there is a way to avoid those heavy interest charges. It can be done for less than R300 extra every
month.
So our payment is now R2 000. The interest rate is still 9% per annum payable monthly (0,75% per
month), and our principal amount borrowed is R190 000. Making this higher repayment amount every
month, how long will it take to pay off the mortgage?
The present value of the stream of payments must be equal to R190 000 (the present value of the
borrowed amount). So we need to solve for n in:
R2 000 x [1  (1 + 0,75%r n ]/0,75% = R190 000
1 (1 + 0,75%) = ( 19000 2 X 00 °' 75% )
lorfl + 0,75%) = log[(l 19000 2 ° X 00 ' 0075 ]
, ,, „, r »,> , r/, 190 000x0,0075,
nxlog(l + 0,75%) = log[(l ^^ ]
n x 0,007472 = 1,2465
n = 166,8 months
= 13,9 years
So the mortgage will be completely repaid in less than 14 years, and you would have made a total
payment of 166,8 x R2 000 = R333 600.
Can you see what is happened? Making regular payments of R2 000 instead of the required Rl 709,48,
you will have saved R76 675,20 (= R410 275,20 R333 600) in interest, and yet you have only paid
an additional amount of R290,52 for 1 66,8 months, or R48 458,74. You surely know by now that the
difference between the additional R48 458,74 that you have paid and the R76 675,20 interest that you
have saved is attributable to, yes, you have got it, compound interest!
10
CHAPTER 4. FINANCE 4.3
Future Value of a Series of Payments mEMCAi
In the same way that when we have a single payment, we can calculate a present value or a future
value  we can also do that when we have a series of payments.
In the above section, we had a few payments, and we wanted to know what they are worth now 
so we calculated present values. But the other possible situation is that we want to look at the future
value of a series of payments.
Maybe you want to save up for a car, which will cost R45 000  and you would like to buy it in 2 years
time. You have a savings account which pays interest of 12% per annum. You need to work out how
much to put into your bank account now, and then again each month for 2 years, until you are ready
to buy the car.
Can you see the difference between this example and the ones at the start of the chapter where we
were only making a single payment into the bank account  whereas now we are making a series of
payments into the same account? This is a sinking fund.
So, using our usual notation, let us write out the answer. Make sure you agree how we come up with
this. Because we are making monthly payments, everything needs to be in months. So let A be the
closing balance you need to buy a car, P is how much you need to pay into the bank account each
month, and i r2 is the monthly interest rate. (Careful  because 12% is the annual interest rate, so we
will need to work out later what the monthly interest rate is!)
A = P(1 + in) 24 + P(l + «i2) 23 + ■ ■ ■ + P(l + J12) 1
Here are some important points to remember when deriving this formula:
1. We are calculating future values, so in this example we use (1 + ir 2 ) n and not (1 + 112)"".
Check back to the start of the chapter if this is not obvious to you by now.
2. If you draw a timeline you will see that the time between the first payment and when you buy
the car is 24 months, which is why we use 24 in the first exponent.
3. Again, looking at the timeline, you can see that the 24" 1 payment is being made one month
before you buy the car  which is why the last exponent is a 1.
4. Always check that you have got the right number of payments in the equation. Check right now
that you agree that there are 24 terms in the formula above.
So, now that we have the right starting point, let us simplify this equation:
A = P[(l + j 12 ) 24 + (l + i 12 ) 23 + ... + (lH 12 ) 1 ]
= P[X 2A + X 23 + ■■■ + X 1 ] using X = (1 + i 12 )
Note that this time X has a positive exponent not a negative exponent, because we are doing future
values. This is not a rule you have to memorise  you can see from the equation what the obvious
choice of X should be.
Let us reorder the terms:
A = PIX 1 + X 2 H h X 24 ] = P . X[l + X + X 2 + h X 23 ]
This is just another sum of a geometric sequence, which as you know can be simplified as:
A = P.X[X n l]/((l + i 12 )l)
= P.X{X n l]/i 12
4.4
CHAPTER 4. FINANCE
So if we want to use our numbers, we know that A =R45 000, n = 24 (because we are looking at
monthly payments, so there are 24 months involved) and i = 12% per annum.
BUT (and it is a big but) we need a monthly interest rate. Do not forget that the trick is to keep the time
periods and the interest rates in the same units  so if we have monthly payments, make sure you use a
monthly interest rate! Using the formula from Grade 1 1, we know that (1 +i) = (1 4 J12) 12 . So we
can show that i 12 = 0,0094888 = 0,94888%.
Therefore,
45 000 = P(l,0094888)[(l,0094888) 24  l]/0,0094888
P = 1662,67
This means you need to invest R166 267 each month into that bank account to be able to pay for your
car in 2 years time.
Exercise 41
1 . You have taken out a mortgage bond for R875 000 to buy a flat. The bond is for 30 years and the
interest rate is 12% per annum payable monthly.
(a) What is the monthly repayment on the bond?
(b) How much interest will be paid in total over the 30 years?
2. How much money must be invested now to obtain regular annuity payments of R5 500 per
month for five years? The money is invested at 11,1% p. a., compounded monthly. (Answer to
the nearest hundred rand.)
A"j More practice f ►) video solutions Cf) or help at www.everythingmaths.co.za
(1.)01e1 (2.) 01e2
4.4 Investments and Loans
By now, you should be well equipped to perform calculations with compound interest. This section
aims to allow you to use these valuable skills to critically analyse investment and loan options that you
will come across in your later life. This way, you will be able to make informed decisions on options
presented to you.
At this stage, you should understand the mathematical theory behind compound interest. However,
the numerical implications of compound interest are often subtle and far from obvious.
Recall the example 'Show Me the Money' in Section 4.3.2. For an extra payment of R29 052 a month,
we could have paid off our loan in less than 14 years instead of 20 years. This provides a good
illustration of the long term effect of compound interest that is often surprising. In the following
section, we'll aim to explain the reason for the drastic reduction in the time it takes to repay the loan.
IS
CHAPTER 4. FINANCE 4.4
Loan Schedules ■ emcak
So far, we have been working out loan repayment amounts by taking all the payments and discounting
them back to the present time. We are not considering the repayments individually. Think about the
time you make a repayment to the bank. There are numerous questions that could be raised: how
much do you still owe them? Since you are paying off the loan, surely you must owe them less money,
but how much less? We know that we'll be paying interest on the money we still owe the bank. When
exactly do we pay interest? How much interest are we paying?
The answer to these questions lie in something called the load schedule.
We will continue to use the earlier example. There is a loan amount of R190 000. We are paying it
off over 20 years at an interest of 9% per annum payable monthly. We worked out that the repayments
should be Rl 709,48.
Consider the first payment of Rl 709,48 one month into the loan. First, we can work out how much
interest we owe the bank at this moment. We borrowed R190 000 a month ago, so we should owe:
M x iia
R190 000 x 0,75%
R1425
We are paying them Rl 425 in interest. We call this the interest component of the repayment. We are
only paying off Rl 709,48— Rl 425 =R284,48 of what we owe! This is called the capital component.
That means we still owe R190 000 R284,48 = R189 715,52. This is called the capital outstanding.
Let's see what happens at the end of the second month. The amount of interest we need to pay is the
interest on the capital outstanding.
M x ii2
R189 715,52 x 0,75%
Rl 422,87
Since we don't owe the bank as much as we did last time, we also owe a little less interest. The capital
component of the repayment is now Rl 709,48— Rl 422,87 = R286,61. The capital outstanding will
be R189 715,52 R286,61 = R189 428,91. This way, we can break each of our repayments down into
an interest part and the part that goes towards paying off the loan.
This is a simple and repetitive process. Table 4.1 is a table showing the breakdown of the first 12
payments. This is called a loan schedule.
Now, let's see the same thing again, but with R2 000 being repaid each year. We expect the numbers
to change. However, how much will they change by? As before, we owe Rl 425 in interest in interest.
After one month. However, we are paying R2 000 this time. That leaves R575 that goes towards paying
off the capital outstanding, reducing it to R189 425. By the end of the second month, the interest owed
is Rl 420,69 (That's R189 425 x i 12 ). Our R2 000 pays for that interest, and reduces the capital amount
owed by R2 000 Rl 420,69 = R579,31. This reduces the amount outstanding to R188 845,69.
Doing the same calculations as before yields a new loan schedule shown in Table 4.2.
The important numbers to notice is the "Capital Component" column. Note that when we are paying
off R2 000 a month as compared to Rl 709,48 a month, this column more than double. In the beginning
of paying off a loan, very little of our money is used to pay off the capital outstanding. Therefore, even
19
4.4
CHAPTER 4. FINANCE
Time
Repayment
Interest Com
Capital Com
Capital Outstand
ponent
ponent
ing
R
190 000,00
1
R
1 709,48
R
1 425,00
R
284,48
R
189 715,52
2
R
1 709,48
R
1 422,87
R
286,61
R
189 428,91
3
R
1 709,48
R
1 420,72
R
288,76
R
189 140,14
4
R
1 709,48
R
1 418,55
R
290,93
R
188 849,21
5
R
1 709,48
R
1 416,37
R
293,11
R
188 556,10
6
R
1 709,48
R
1414,17
R
295,31
R
188 260,79
7
R
1 709,48
R
1411,96
R
297,52
R
187 963,27
8
R
1 709,48
R
1 409,72
R
299,76
R
187 663,51
9
R
1 709,48
R
1 407,48
R
302,00
R
187 361,51
10
R
1 709,48
R
1 405,21
R
304,27
R
187 057,24
11
R
1 709,48
R
1 402,93
R
306,55
R
186 750,69
12
R
1 709,48
R
1 400,63
R
308,85
R
186 441,84
Table 4.1 : A loan schedule with repayments of Rl 709,48 per month.
Time
Repayment
Interest Com
Capital Com
Capital Outstand
ponent
ponent
ing
R
190 000,00
1
R
2 000,00
R
1 425,00
R
575,00
R
189 425,00
2
R
2 000,00
R
1 420,69
R
579,31
R
188 845,69
3
R
2 000,00
R
1 416,34
R
583,66
R
188 262,03
4
R
2 000,00
R
1411,97
R
588,03
R
187 674,00
5
R
2 000,00
R
1 407,55
R
592,45
R
187 081,55
6
R
2 000,00
R
1 403,11
R
596,89
R
186 484,66
7
R
2 000,00
R
1 398,63
R
601,37
R
185 883,30
8
R
2 000,00
R
1 394,12
R
605,88
R
185 277,42
9
R
2 000,00
R
1 389,58
R
610,42
R
184 667,00
10
R
2 000,00
R
1 385,00
R
615,00
R
184 052,00
11
R
2 000,00
R
1 380,39
R
619,61
R
183 432,39
12
R
2 000,00
R
1 375,74
R
624,26
R
182 808,14
Table 4.2: A loan schedule with repayments of R2 000 per month.
50
CHAPTER 4. FINANCE 4.4
a small increase in repayment amounts can significantly increase the speed at which we are paying off
the capital.
What's more, look at the amount we are still owing after one year (i.e. at time 12). When we were
paying Rl 709,48 a month, we still owe R186 441,84. However, if we increase the repayments to
R2 000 a month, the amount outstanding decreases by over R3 000 to R182 808,14. This means we
would have paid off over R7 000 in our first year instead of less than R4 000. This increased speed at
which we are paying off the capital portion of the loan is what allows us to pay off the whole loan
in around 14 years instead of the original 20. Note however, the effect of paying R2 000 instead of
Rl 709,48 is more significant in the beginning of the loan than near the end of the loan.
It is noted that in this instance, by paying slightly more than what the bank would ask you to pay, you
can pay off a loan a lot quicker. The natural question to ask here is: why are banks asking us to pay the
lower amount for much longer then? Are they trying to cheat us out of our money?
There is no simple answer to this. Banks provide a service to us in return for a fee, so they are out to
make a profit. However, they need to be careful not to cheat their customers for fear that they'll simply
use another bank. The central issue here is one of scale. For us, the changes involved appear big. We
are paying off our loan 6 years earlier by paying just a bit more a month. To a bank, however, it doesn't
matter much either way. In all likelihood, it doesn't affect their profit margins one bit!
Remember that the bank calculates repayment amounts using the same methods as we've been learn
ing. They decide on the correct repayment amounts for a given interest rate and set of terms. Smaller
repayment amounts will make the bank more money, because it will take you longer to pay off the
loan and more interest will accumulate. Larger repayment amounts mean that you will pay off the loan
faster, so you will accumulate less interest i.e. the bank will make less money off of you. It's a simple
matter of less money now or more money later. Banks generally use a 20 year repayment period by
default.
Learning about financial mathematics enables you to duplicate these calculations for yourself. This
way, you can decide what's best for you. You can decide how much you want to repay each month
and you'll know of its effects. A bank wouldn't care much either way, so you should pick something
that suits you.
Example 4: Monthly Payments
QUESTION
Stefan and Mama want to buy a house that costs R 1 200 000. Their parents offer to put down
a 20% payment towards the cost of the house. They need to get a mortgage for the balance.
What are their monthly repayments if the term of the home loan is 30 years and the interest is
7,5%, compounded monthly?
SOLUTION
Step 1 : Determine how much money they need to borrow
Rl 200 00R240 000 =R960 000
Step 2 : Determine how to approach the problem
Use the formula:
p _ x[l(l + i)"]
Where
P =R960 000
4.4
CHAPTER 4. FINANCE
n = 30 x 12 = 360 months
4 = 0,075^12 = 0,00625
Step 3 : Solve the problem
R960 000
x[l(l + 0,00625)~ 360 ]
0,00625
x(143,0176273)
R6 712,46
Step 4 : Write the final answer
The monthly repayments =R6 712,46
Exercise 42
1. A property costs Rl 800 000. Calculate the monthly repayments if the interest rate is 14% p. a.
compounded monthly and the loan must be paid off in 20 years time.
2. A loan of R 4 200 is to be returned in two equal annual instalments. If the rate of interest of 10%
per annum, compounded annually, calculate the amount of each instalment.
f/Vj More practice Crj video solutions C ?J or help at www.everythingmaths.co.za
(1.)01e3 (2.)01e4
Calculating Capital Outstanding
EMCAL
As defined in Section 4.4.1, capital outstanding is the amount we still owe the people we borrowed
money from at a given moment in time. We also saw how we can calculate this using loan schedules.
However, there is a significant disadvantage to this method: it is very time consuming. For example, in
order to calculate how much capital is still outstanding at time 12 using the loan schedule, we'll have
to first calculate how much capital is outstanding at time 1 through to 11 as well. This is already quite
a bit more work than we'd like to do. Can you imagine calculating the amount outstanding after 10
years (time 120)?
Fortunately, there is an easier method. However, it is not immediately clear why this works, so let's
take some time to examine the concept.
Prospective Method for Capital Outstanding
Let's say that after a certain number of years, just after we made a repayment, we still owe amount Y.
What do we know about Yl We know that using the loan schedule, we can calculate what it equals
to, but that is a lot of repetitive work. We also know that Y is the amount that we are still going to pay
CHAPTER 4. FINANCE
4.5
off. In other words, all the repayments we are still going to make in the future will exactly pay off Y.
This is true because in the end, after all the repayments, we won't be owing anything.
Therefore, the present value of all outstanding future payments equal the present amount outstanding.
This is the prospective method for calculating capital outstanding.
Let's return to a previous example. Recall the case where we were trying to repay a loan of R200 000
over 20 years. A RIO 000 deposit was put down, so the amount being payed off was R190 000. At an
interest rate of 9% compounded monthly, the monthly repayment was Rl 709,48. In Table 4.1, we can
see that after 12 months, the amount outstanding was R186 441,84. Let's try to work this out using the
the prospective method.
After time 12, there are still 19 x 12 = 228 repayments left of Rl 709,48 each. The present value is:
228
0,75%
Y
Rl 709,48 x
R186 441,92
1  1,0075'
0,0075
Oops! This seems to be almost right, but not quite. We should have got R186 441,84. We are 8
cents out. However, this is in fact not a mistake. Remember that when we worked out the monthly
repayments, we rounded to the nearest cents and arrived at Rl 709,48. This was because one cannot
make a payment for a fraction of a cent. Therefore, the rounding off error was carried through. That's
why the two figures don't match exactly. In financial mathematics, this is largely unavoidable.
4.5 Formula Sheet
As an easy reference, here are the key formulae that we derived and used during this chapter. While
memorising them is nice (there are not many), it is the application that is useful. Financial experts are
not paid a salary in order to recite formulae, they are paid a salary to use the right methods to solve
financial problems.
Definitions
EMCAN
P Principal (the amount of money at the starting point of the calculation)
i interest rate, normally the effective rate per annum
n period for which the investment is made
iT the interest rate paid T times per annum, i.e. iT ■■
Nominal Interest Rate
r,:i
4.5
CHAPTER 4. FINANCE
Equations
EMCAO
Present Value  simple
Future Value  simple
Solve for i
Solve for n
P(l + i.n)
Tip
Always keep the interest
and the time period in
the same units of time
(e.g. both in years, or
both in months etc.).
Present Value  compound
Future Value  compound
Solve for i
Solve for n
P(l + *)"
Chapter 4
End of Chapter Exercises
1. Thabo is about to invest his R8 500 bonus in a special banking product which will
pay 1% per annum for 1 month, then 2% per annum for the next 2 months, then 3%
per annum for the next 3 months, 4% per annum for the next 4 months, and 0% for
the rest of the year. The bank is going to charge him R100 to set up the account. How
much can he expect to get back at the end of the period?
2. A special bank account pays simple interest of 8% per annum. Calculate the opening
balance required to generate a closing balance of R5 000 after 2 years.
3. A different bank account pays compound interest of 8% per annum. Calculate the
opening balance required to generate a closing balance of R5 000 after 2 years.
4. Which of the two answers above is lower, and why?
5. 7 months after an initial deposit, the value of a bank account which pays compound
interest of 7,5% per annum is R3 650,81. What was the value of the initial deposit?
6. Thabani and Lungelo are both using UKZN Bank for their saving. Suppose Lungelo
makes a deposit of X today at interest rate of i for six years. Thabani makes a deposit
of 3X at an interest rate of 0,05%. Thabani made his deposit 3 years after Lungelo
made his first deposit. If after 6 years, their investments are equal, calculate the value
of i and find X. If the sum of their investment is R20 000, use the value of X to find
out how much Thabani earned in 6 years.
7. Sipho invests R500 at an interest rate of log(l,12) for 5 years. Themba, Sipho's sister
invested R200 at interest rate i for 1 years on the same date that her brother made his
first deposit. If after 5 years, Themba's accumulation equals Sipho's, find the interest
rate i and find out whether Themba will be able to buy her favourite cell phone after
10 years which costs R2 000.
8. Calculate the real cost of a loan of R10 000 for 5 years at 5% capitalised monthly.
Repeat this for the case where it is capitalised half yearly i.e. Every 6 months.
9. Determine how long, in years, it will take for the value of a motor vehicle to decrease
to 25% of its original value if the rate of depreciation, based on the reducingbalance
method, is 21% per annum.
51
CHAPTER 4. FINANCE 4.5
A" 1 ) More practice (Vj video solutions (?) or help at www.everythingmaths.co.z
(1.)01e5 (2.)01e6 (3.) 01e7 (4.) 01e8 (5.) 01e9 (6.) 01ea
(7.)01eb (8.)01ec (9.) 01 ed
Factorising Cubic
Polynomials
5. 1 Introduction
In Grades 10 and 1 1, you learnt how to solve different types of equations. Most of the solutions, relied
on being able to factorise some expression and the factorisation of quadratics was studied in detail.
This chapter focuses on the factorisation of cubic polynomials, that is expressions with the highest
power equal to 3.
See introductory video: VMgmo at www.everythingmaths.co.za
5.2 The Factor Theorem
The factor theorem describes the relationship between the root of a polynomial and a factor of the
polynomial.
DEFINITION: Factor Theorem
For any polynomial, f(x), for al
of f(x). Or, more concisely:
values of a which satisfy f(a) =
(x
 a) is a factor
fix) = {xa)q(x)
where q(x) is a polynomial.
In other words: If the remainder when dividing f(x) by (x — a)
is a factor of f(x).
So if /(£) = 0, then (ax + b) is a factor of f{x).
is
zero
then
[x — a)
Example 1: Factor Theorem
56
CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 5.2
QUESTION
Use the Factor Theorem to determine whether y — 1 is a factor of f(y) = 2y 4 + 3y 2 — by + 7.
SOLUTION
Step 1 : Determine how to approach the problem
In order for y — 1 to be a factor, /(l) must be 0.
Step 2 : Calculate /(l)
f(y) = 2y i + 3y 2 by + 7
../(l) = 2(l) 4 + 3(l) 2 5(l) + 7
= 2+35+7
= 7
Step 3 : Conclusion
Since /(l) ^ 0, y  1 is not a factor of /(j/) = 2j/ 4 + 3j/ 2  5?/ + 7.
Example 2: Factor Theorem
QUESTION
Using the Factor Theorem, verify that y + 4 is a factor of g(y) = by + 16y — \by + 8y + 16
SOLUTION
Step 1 : Determine how to approach the problem
In order for y + 4 to be a factor, g(— 4) must be 0.
Step 2 : Calculate /(l)
g(y) = by" + 16j/ 3  lby 2 + Sy + 16
R..g(A) = 5(4) 4 + 16(4) 3 15(4) 2 +8(4) + 16
= 5(256) + 16(64) 15(16) + 8(4) + 16
= 1280  1024  240 32+16
=
Step 3 : Conclusion
5.3
CHAPTER 5. FACTORISING CUBIC POLYNOMIALS
Since ff(4) = 0, y + 4 is a factor of g(y) = 5?/ 4 + 16y 3  15y 2 + 8j/ + 16.
5.3 Factorisation of Cubic
Polynomials
A cubic polynomial is a polynomial of the form
ax + fox + ex + d
where a is nonzero. We have seen in Grade 10 that the sum and difference of cubes is factorised as
follows:
(x + y)(x 2 xy + y 2 ) = x 3 + y 3
and
(x  y)(x 2 + xy + y 2 ) = x 3  y 3
We also saw that the quadratic factor does not have real roots.
There are many methods of factorising a cubic polynomial. The general method is similar to that used
to factorise quadratic equations. If you have a cubic polynomial of the form:
/(x) = ax + fox + ex + d
then in an ideal world you would get factors of the form:
(Ax + B)(Cx + D)(Ex + F). (5.1)
But sometimes you will get factors of the form:
{Ax + B)(Cx 2 +Ex + D)
We will deal with simplest case first. When a = 1, then A = C = E = 1, and you only have to
determine B, D and F. For example, find the factors of:
In this case we have
a = 1
fo = 2
c = 5
d = 6
The factors will have the general form shown in (5.1), with A = C = E = 1. We can then use values
for a, fo, c and d to determine values for B, D and F. We can rewrite (5.1) with A = C = E = 1 as:
If we multiply this out we get:
{x + B)(x + D)(x + F)
We can therefore write:
(x + B){x + D){x + F).
(x + B)(x 2 + Dx + Fx + DF)
x 3 + Dx 2 + Fx 2 + Bx 2 + DFx + BDx + BFx + BDF
x 3 + (D + F + B)x 2 + {DF + BD + BF)x + BDF
fo =
2 = D + F + B
c =
5 = DF + BD + BF
d =
6 = BDF.
(5.2)
(5.3)
(5.4)
CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 5.3
This is a set of three equations in three unknowns. However, we know that B, D and F are factors of
6 because BDF = 6. Therefore we can use a trial and error method to find B, D and F.
This can become a very tedious method, therefore the Factor Theorem can be used to find the factors
of cubic polynomials.
Example 3: Factorisation of Cubic Polynomials
QUESTION
Factorise f(x) = x 3 + x 2 — 9x — 9 into three linear factors.
SOLUTION
Step I : By trial and error using the factor theorem to find a factor
Try
/(l) = (l) 3 + (l) 2  9(1) 9 = 1 + 199 = 16
Therefore (x — 1) is not a factor
Try
/(l) = (1) 3 + (1) 2 9(1)9 = 1 + 1 + 99 =
Thus (x + 1) is a factor, because /(— 1) = 0.
Now divide f(x) by (x + 1) using division by inspection:
Write x 3 + x 2 9x9 = (x + 1)( )
The first term in the second bracket must be a; 2 to give x 3 if one works backwards.
The last term in the second bracket must be —9 because +1 x — 9 = —9.
So we have x 3 + x 2  9x  9 = (x + l)(x 2 +?x  9).
Now, we must find the coefficient of the middle term (x).
(+l)(x 2 ) gives the x 2 in the original polynomial. So, the coefficient of the xterm
must be 0.
So/(x) = (x + l)(x 2 9).
Step 2 : Factorise fully
x 2 — 9 can be further factorised to (x — 3)(x + 3),
and we are now left with /(x) = (x + l)(x — 3)(x + 3)
In general, to factorise a cubic polynomial, you find one factor by trial and error. Use the factor theorem
to confirm that the guess is a root. Then divide the cubic polynomial by the factor to obtain a quadratic.
Once you have the quadratic, you can apply the standard methods to factorise the quadratic.
For example the factors of x 3 — 2x 2 — 5x + 6 can be found as follows: There are three factors which
we can write as
(x — a)(x — b)(x — c).
5.3 CHAPTER 5. FACTORISING CUBIC POLYNOMIALS
Example 4: Factorisation of Cubic Polynomials
QUESTION
Use the Factor Theorem to factorise
SOLUTION
Step 1 : Find one factor using the Factor Theorem
Try
/(l) = (l) 3  2(1) 2  5(1) + 6 = 125 + 6 =
Therefore (x — 1) is a factor.
Step 2 : Division by inspection
x 3 2x 2 5x + 6 = (x 1)( )
The first term in the second bracket must be x 2 to give x 3 if one works backwards.
The last term in the second bracket must be —6 because — 1 x — 6 = +6.
So we have x 3  2x 2  5x + 6 = (x  l)(x 2 +?x  6).
Now, we must find the coefficient of the middle term (x).
(— l)(x 2 ) gives —x 2 . So, the coefficient of the xterm must be —1.
So/(x) = (x \){x 2 x6).
Step 3 : Factorise fully
x 2 — x — 6 can be further factorised to (x — 3)(x + 2),
and we are now left with x 3 — 2x 2 — bx + 6 = (x — l)(x — 3)(x + 2)
Exercise 51
1 . Find the remainder when 4r 3 — Ax 2 + x — 5 is divided by (x + 1).
2. Use the factor theorem to factorise x 3 — 3x 2 + 4 completely.
3. f(x) = 2x 3 + x 2 5x + 2
(a) Find /(l).
(b) Factorise f(x) completely
4. Use the Factor Theorem to determine all the factors of the following expression:
x" + x — YJx + 15
5. Complete: If f(x) is a polynomial and p is a number such that f(p) = 0, then (x — p) is..
60
CHAPTER 5. FACTOR1SING CUBIC POLYNOMIALS
5.4
More practice (►) video solutions (?) or help at www.everythingmaths.co.:
(1.)01eh (2.)01ei (3.) Olej (4.) 01ek (5.) Olem
5.4 Solving Cubic Equations
Once you know how to factorise cubic polynomials, it is also easy to solve cubic equations of the kind
ax + bx" + ex + d =
Example 5: Solution of Cubic Equations
QUESTION
Solve
6x 3  5x 2  17x + 6 = 0.
SOLUTION
Step 1
: Find one factor using the Factor Theorem
Try
/(l) =6(1) 3 5(1) 2 17(1) + 6 = 6 5 17 + 6 = 
10
Therefore (x — 1) is NOT a factor.
Try
/(2) = 6(2) 3  5(2) 2  17(2) + 6 = 48  20  34 + 6 =
=
Therefore (x — 2) IS a factor.
Step 2
: Division by inspection
6x 3  5x 2  17x + 6 = (x  2){ )
The first term in the second bracket must be 6a; 2 to give 6x
wards.
The last term in the second bracket must be —3 because —2
So we have 6x 3  5x 2  Ylx + 6 = (x  2){6x 2 +?x  3).
Now, we must find the coefficient of the middle term (x).
(— 2)(6x 2 ) gives — 12x 2 . So, the coefficient of the zterm mi
So, 6a; 3  5x 2  17s + 6 = (x  2)(6ir 2 + 7x  3).
3 if one
x 3 =
st be 7.
works back
+6.
(il
5.4 CHAPTER 5. FACTORISING CUBIC POLYNOMIALS
Step 3 : Factorise fully
6x 2 + 7x  3 can be further factorised to (2x + 3)(3aj  1),
and we are now left with 6x 3  5x 2  17cc + 6 = (a;  2)(2s + 3)(3x  1)
Step 4 : Solve the equation
5x 2  YJx + 6 =
(»2)(2w + 3)(3x 1)
2;i; 3
3 2
Sometimes it is not possible to factorise the trinomial ("second bracket"). This is when the quadratic
formula
b ± sJW  iac
2a
can be used to solve the cubic equation fully.
For example:
Example 6: Solution of Cubic Equations
QUESTION
Solve for x: x 3  2x 2  Gx + 4 = 0.
SOLUTION
Step 1 : Find one factor using the Factor Theorem
Try
/(1) =
(l) 3 
2(1) 2
6(1)
Therefore (x
 1) is
NOT a
factor.
Try
/(2) = (2) 3  2(2) 2  6(2) + 4 = 88 12 + 4 = 8
Therefore (x — 2) is NOT a factor.
/(2) = (2) 3  2(2) 2  6(2) + 4 = 8 8 + 12 + 4 =
Therefore (x + 2) IS a factor.
Step 2 : Division by inspection
x 3 2x 2 6x + 4 = (x + 2)( )
The first term in the second bracket must be x 2 to give x 3 .
The last term in the second bracket must be 2 because 2x2 = +4.
So we have x 3  2x 2  6x + 4 = (x + 2)(x 2 +?x + 2).
Now, we must find the coefficient of the middle term (x).
(2)(x 2 ) gives 2x 2 . So, the coefficient of the xterm must be —4. (2x 2 — Ax 2
62
CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 5.4
2x 2 )
So x 3  2x 2  6x + A = (x + 2)(x 2  Ax + 2).
x 2 — Ax + 2 cannot be factorised any further and we are now left with
(x + 2)(x 2 Ax + 2) =
Step 3 : Solve the equation
{x + 2)(x 2  Ax + 2) =
(x + 2) = or (x 2 4x + 2) =
Step 4 : Apply the quadratic formula for the second bracket
Always write down the formula first and then substitute the values of a, b and c.
X
b ± Vb 2  Aac
2a
(4)±V(4) 2 
4(1)(2)
2(1)
A±V8
2
= 2±^2
Step 5 : Final solutions
x = — 2 or x =
2±V2
Chapter 5
End of Chapter Exercises
1 . Solve for x: x 3 + x 2 — 5x + 3 =
2. Solve for y: y 3  3y 2  l&y  12 =
3. Solve for m: m 3 — m 2 — Am — 4 =
4. Solve for x: x 3 — x 2 = 3(3x + 2) Tip: Remove brackets and write as an equation
equal to zero.
5. Solve for x if 2x 3  3x 2  8x = 3
6. Solve for x: 16(x + 1) = x 2 (x + 1)
7. (a) Show that x  2 is a factor of 3x 3  llx 2 + 12x  4
(b) Hence, by factorising completely, solve the equation
3x 3  llx 2 + 12x  4 =
8. 2x 3  x 2  2x + 2 = Q(x).(2x 1) + i? for all values of x. What is the value of ffl.
9. (a) Use the factor theorem to solve the following equation for m:
8m 3 + 7m 2  17m + 2 =
(b) Hence, or otherwise, solve for x:
(53
5.4 CHAPTER 5. FACTORISING CUBIC POLYNOMIALS
10. A challenge:
Determine the values of p for which the function
fix) = 3p — (3p — 7)x + 5x — 3
leaves a remainder of 9 when it is divided by (x — p).
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(1.)01en (2.)01ep (3.) 01eq (4.) 01er (5.)01es (6.) 01et
(7.)01eu (8.) 01ev (9.) 01ew (10.) 01ex
()i
Functions and Graphs
6. 1 Introduction
In Grades 10 and 1 1 you learned about linear functions and quadratic functions, as well as the hyper
bolic functions and exponential functions and many more. In Grade 12 you are expected to demon
strate the ability to work with various types of functions and relations including inverses. In particular,
we will look at the graphs of the inverses of:
y = ax + q
2
y = ax
y = ax; a >
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6.2 Definition of a Function
A function is a relation for which there is only one value of y corresponding to any value of x. We
sometimes write y = fix), which is notation meaning 'y is a function of x'. This definition makes
complete sense when compared to our real world examples — each person has only one height, so
height is a function of people; on each day, in a specific town, there is only one average temperature.
However, some very common mathematical constructions are not functions. For example, consider the
relation x 2 + y 2 = 4. This relation describes a circle of radius 2 centred at the origin, as in Figure 6.1 .
If we let x = 0, we see that y 2 = 4 and thus either y
which are possible for the same x value, the relation x 2
2 or y = —2. Since there are two y values
 y 2 = 4 is not the graph a function.
There is a simple test to check if a relation is a function, by looking at its graph. This test is called the
vertical line test. If it is possible to draw any vertical line (a line of constant x) which crosses the graph
of the relation more than once, then the relation is not a function. If more than one intersection point
exists, then the intersections correspond to multiple values of y lor a single value of x.
We can see this with our previous example of the circle by looking at its graph again in Figure 6.1.
We see that we can draw a vertical line, for example the dotted line in the drawing, which cuts the
circle more than once. Therefore this is not a function.
Exercise 67
1 . State whether each of the following equations are functions or not:
(a) x + y = 4
(b) y = f
(c) y = T
65
6.3
CHAPTER 6. FUNCTIONS AND GRAPHS
Figure 6.1: Graph of x 2
(d)
If
2. The table gives the average per capita income, d, in a region of the country as a function of u,
the percentage unemployed. Write down the equation to show that income is a function of the
percent unemployed.
u
1
2
3
4
d
22500
22000
21500
21000
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(1.)0ley (2.)01ez
6.3 Notation Used for Functions
In Grade 1 you were introduced to the notation used to name a function. In a function y = f(x), y is
called the dependent variable, because the value of y depends on what you choose as x. We say x is
the independent variable, since we can choose x to be any number. Similarly if g(t) = 2t + 1, then t
is the independent variable and g is the function name. If f(x) = 3x — 5 and you are ask to determine
/(3), then you have to work out the value for f(x) when x = 3. For example,
/(3)
3x —
3(3)
4
(i(i
CHAPTER 6. FUNCTIONS AND GRAPHS 6.4
6.4 Graphs of Inverse Functions
In earlier grades, you studied various types of functions and understood the effect of various parameters
in the general equation. In this section, we will consider inverse functions.
An inverse function is a function which does the reverse of a given function. More formally, if / is a
function with domain X, then / _1 is its inverse function if and only if for every ielwe have:
r 1 u(x)) = x (6.D
A simple way to think about this is that a function, say y = f(x), gives you a yvalue if you substitute an
ivalue into f(x). The inverse function tells you tells you which xvalue was used to get a particular
j/value when you substitute the yvalue into f~ 1 (y). There are some things which can complicate
this  for example, for the sin function there are many xvalues that give you a peak as the function
oscillates. This means that the inverse of the sin function would be tricky to define because if you
substitute the peak j/value into it you won't know which of the ivalues was used to get the peak.
y = f(x) we have a function
g/i = f{xi) we substitute a specific xvalue into the function to get a specific j/value
consider the inverse function
x = f' 1 {y)
x = f~ (y) substituting the specific yvalue into the inverse should return the specific rcvalue
= F\yi)
= X!
This works both ways, if we don't have any complications like in the case of the sin function, so we
can write:
r 1 (f(x)) = f(r 1 (x)) = x (6.2)
(x  2)
For example, if the function x — > 3x + 2 is given, then its inverse function is x — >  —  — . This is
usually written as:
/ : x^3x + 2 (6.3)
r 1 ■ »^ (6.4)
The superscript " — 1" is not an exponent.
If a function / has an inverse then / is said to be invertible.
If / is a realvalued function, then for / to have a valid inverse, it must pass the horizontal line test,
that is a horizontal line y = k placed anywhere on the graph of / must pass through / exactly once
for all real k.
It is possible to work around this condition, by defining a multivalued function as an inverse.
If one represents the function / graphically in a zycoordinate system, the inverse function of the
equation of a straight line, f 1 , is the reflection of the graph of / across the line y = x.
Algebraically, one computes the inverse function of / by solving the equation
y = f(x)
for x, and then exchanging y and x to get
y = /^M
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6.4
CHAPTER 6. FUNCTIONS AND GRAPHS
Inverse Function ofy = ax + q
EMCAX
The inverse function of y = ax + q is determined by solving for x as:
(6.5)
(6.6)
(6.7)
(6.8)
Therefore the inverse of y ■■
The inverse function of a straight line is also a straight line, except for the case where the straight line
is a perfectly horizontal line, in which case the inverse is undefined.
For example, the straight line equation given by y = 2x — 3 has as inverse the function, y — x + §.
The graphs of these functions are shown in Figure 6.2. It can be seen that the two graphs are reflections
of each other across the line y = x.
y

ax + q
ax
=
yq
X
=
yq
a
=
i q
v
a a
ax + q\iy = X
_ £
a
r\x)
Figure 6.2: The graphs of the function f(x) = 2x — 3 and its inverse / i (x) = \x+ ^. The line y ■■
is shown as a dashed line.
Domain and Range
We have seen that the domain of a function of the form y = ax + q is {x : x e M} and the range is
{y : y 6 M}. Since the inverse function of a straight line is also a straight line, the inverse function will
have the same domain and range as the original function.
Intercepts
The general form of the inverse function of the form y = ax + q is y = x — .
(i.s
CHAPTER 6. FUNCTIONS AND GRAPHS 6.4
By setting i = Owe have that the ^intercept is y int = — J. Similarly, by setting y = we have that
the xintercept is Xmt = 1
It is interesting to note that if f(x) = ax + q, then f 1 {x) = x — 2 and the j/intercept of f(x) is the
zintercept of / _1 (x) and the xintercept of f(x) is the j/intercept of f~ L (x).
Exercise 62
1 . Given f(x) = 2x  3, find f^{x)
2. Consider the function f(x) = 3x — 7.
(a) Is the relation a function?
(b) If it is a function, identify the domain and range.
3. Sketch the graph of the function f(x) = 3x — 1 and its inverse on the same set of axes.
4. The inverse of a function is f^ 1 (x) = 2x — 4, what is the function /(a;)?
(f^j More practice Crj video solutions Cf) or help at www.everythingmaths.co.za
(1.)01fO (2.) 01 f1 (3.)01f2 (4.) 01 f3
Inverse Function ofy = ax 2 wemcay
The inverse relation, possibly a function, of y = ax 2 is determined by solving for x as:
y = ax (6.9)
x 2 ¥■ (6.10)
±J (6.11)
(i'l
6.4
CHAPTER 6. FUNCTIONS AND GRAPHS
f(x)=x 2
r\*) = sfc
r i (x) = ^
Figure 6.3: The function f(x) = x 2 and its inverse / L (x) = ±^/x. The line y = x is shown as a
clashed line.
We see that the inverse relation of y
is not a function because it fails the vertical line test. If
we draw a vertical line through the graph of f~ 1 (x) = ±^/x, the line intersects the graph more than
once. There has to be a restriction on the domain of a parabola for the inverse to also be a function.
Consider the function f(x) = —x 2 + 9. The inverse of / can be found by writing f(y) = x. Then
x = —y + 9
y = 9 — x
y = ±\/9x
If we restrict the domain of f(x) to be x > 0, then \/9 — x is a function. If the restriction on the
domain of / is x < then —\/9 — x would be a function, inverse to /.
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Exercise 63
1 . The graph of / Ms shown. Find the equation of /, given that the graph of / is a parabola. (Do
not simplify your answer)
CHAPTER 6. FUNCTIONS AND GRAPHS 6.4
2. f(x) = 2x 2 .
(a) Draw the graph of / and state its domain and range.
(b) Find / _1 and, if it exists, state the domain and range.
(c) What must the domain of / be, so that / _1 is a function ?
3. Sketch the graph of x = — y/10 — y 2 . Label a point on the graph other than the intercepts with
the axes.
4. (a) Sketch the graph of y = x 2 labelling a point other than the origin on your graph.
(b) Find the equation of the inverse of the above graph in the form y = . . .
(c) Now sketch the graph of y = </x.
(d) The tangent to the graph of y = sfx at the point A(9; 3) intersects the zaxis at B. Find the
equation of this tangent and hence or otherwise prove that the yaxis bisects the straight
line AB.
5. Given: g(x) = — 1 + </x, find the inverse of g(x) in the form g^ 1 {x) = . . .
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Inverse Function ofy = a x wemcaz
The inverse function of y = a x is determined by solving for x as follows:
y = a (6.12)
log(j/) = log(a*) (6.13)
= xlog(a) (6.14)
log(y)
log(a)
The inverse of y = 1CT is x = log(y) Therefore, if f(x) = lO 1 ', then /(x) _1 = log(x).
(6.15)
71
6.4
CHAPTER 6. FUNCTIONS AND GRAPHS
fix) = 10*
r i (x) = io g (x)
Figure 6.4: The function f(x) = 1(F and its inverse / 1 (x) = log(x). The line y = x is shown as a
clashed line.
The exponential function and the logarithmic function are inverses of each other; the graph of the one
is the graph of the other, reflected in the line y = x. The domain of the function is equal to the range
of the inverse. The range of the function is equal to the domain of the inverse.
Exercise 64
1 . Given that f(x) = (\) x , sketch the graphs of / and / x on the same system of axes indicating
a point on each graph (other than the intercepts) and showing clearly which is / and which is
r 1 .
2. Given that /(x) = 4 _x ,
(a) Sketch the graphs of /and f~ l on the same system of axes indicating a point on each graph
(other than the intercepts) and showing clearly which is / and which is / _1 .
(b) Write / _1 in the form y = . . .
3. Given g(x) = — 1 + \[x, find the inverse of g(x) in the form g^ 1 (x) = . . .
4. (a) Sketch the graph of y = x 2 , labelling a point other than the origin on your graph.
(b) Find the equation of the inverse of the above graph in the form y = . . .
(c) Now, sketch y = *Jx.
(d) The tangent to the graph of y = *Jx at the point A(9; 3) intersects the xaxis at B. Find the
equation of this tangent, and hence, or otherwise, prove that the j/axis bisects the straight
line AB.
72
CHAPTER 6. FUNCTIONS AND GRAPHS
6.4
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Chapter 6
End of Chapter Exercises
1 . Sketch the graph of x = — ^/10 — y 2 . Is this graph a function? Verify your answer.
2 /(*) = ^=,
x — 5
(a) determine the j/intercept of f(x)
(b) determine x if f(x) = — 1.
3. Below, you are given three graphs and five equations.
Graph 1 Graph 2 Graph 3
I MP x
(a) j/ = log 3 x
(b) 3/ =  log 3 x
(c) 2/ = log 3 (x)
(d) y = 3"
(e) y = T
Write the equation that best matches each graph.
4. Given g(x) = — 1 + y^< find the inverse of g(x) in the form g^ 1 (x) = . . .
5. Consider the equation h(x) = 3 X
(a) Write down the inverse in the form /i _1 (x) = . . .
(b) Sketch the graphs of h(x) and h~ 1 (x) on the same set of axes, labelling the
intercepts with the axes.
(c) For which values of x is h^ 1 (x) undefined ?
6. (a) Sketch the graph of y = 2x 2 + 1, labelling a point other than the origin on your
graph.
6.4 CHAPTER 6. FUNCTIONS AND GRAPHS
(b) Find the equation of the inverse of the above graph in the form y = . . .
(c) Now, sketch y = s/x.
(d) The tangent to the graph of y = ^/x at the point A(9; 3) intersects the xaxis at
B. Find the equation of this tangent, and hence, or otherwise, prove that the
yaxis bisects the straight line AB.
More practice (►) video solutions (0 J or help at www.everythingmaths.co.za
(1.)01fb (2.)01fc (3.) Olfd (4.)01fe (5.) 01ff (6.) Olfg
7i
Differential Calculus
7.1 Introduction
Calculus is one of the central branches of mathematics and was developed from algebra and geometry.
Calculus is built on the concept of limits, which will be discussed in this chapter. Calculus consists
of two complementary ideas: differential calculus and integral calculus. We will only be dealing with
differential calculus in this text and will explore how it can be used to address mathematical problems
like optimisation and finding rates of change.
© See introductory video: VMgxl at www.everythingmaths.co.za
7.2 Limits
The Tale of Achilles and the Tortoise
EMCBC
One of Zeno's paradoxes can be summarised as:
Achilles and a tortoise agree to a race, but the tortoise is unhappy because Achilles is very
fast. So, the tortoise asks Achilles for a headstart. Achilles agrees to give the tortoise a
1000m head start. Does Achilles overtake the tortoise?
We know how to solve this problem. We start by writing:
xa = 11 At (7.1)
x t = 1000m + v t t (7.2)
FACT
Zeno (circa 490 BC 
circa 430 BC) was a pre
Socratic Greek philoso
pher of southern Italy
who is famous for his
paradoxes.
where
xa distance covered by Achilles
va Achilles' speed
t time taken by Achilles to overtake tortoise
x t distance covered by the tortoise
Vt the tortoise's speed
If we assume that Achilles runs at 2m ■ s _1 and the tortoise runs at 0,25 1
overtake the tortoise when both of them have covered the same distance,
overtakes the tortoise at a time calculated as:
is 1 then Achilles will
This means that Achilles
75
7.2 CHAPTER 7. DIFFERENTIAL CALCULUS
xa = xt (7.3)
VAt = 1000 + v t t (7.4)
(2ms _1 )i = 1000m + (0,25 ms _1 )t (7.5)
(2m s _1 0,25m s _1 )i = 1000m (7.6)
t = ™^ (7.7,
l4m s 1
1000 m
(4) (1000)
(7.8)
(7.9)
= <%>. (7,0,
= 571^s (7.11)
However, Zeno (the Greek philosopher who thought up this problem) looked at it as follows: Achilles
takes
1000 m
t = r = 500 s
2m ■ s 1
to travel the 1 000 m head start that the tortoise had. However, in these 500 s, the tortoise has travelled
a further
x = (500s)(0,25ms _1 ) = 125 m.
Achilles then takes another
125 m
t = ~ r = 62 ' 5s
2 m s 1
to travel the 125 m. In these 62,5 s, the tortoise travels a further
x = (62,5s)(0,25ms _1 ) = 15,625 m.
Zeno saw that Achilles would always get closer but wouldn't actually overtake the tortoise.
Sequences, Series and Functions m emcbd
So what does Zeno, Achilles and the tortoise have to do with calculus?
Well, in Grades 10 and 1 1 you studied sequences. For the sequence
oi 2  3  4 
'2'3'4'5'
which is defined by the expression
On = 1
n
the terms get closer to 1 as n gets larger. Similarly, for the sequence
' 2'3'4' 5'
which is defined by the expression
1
On = —
11
the terms get closer to as n gets larger. We have also seen that the infinite geometric series can have
a finite total. The infinite geometric series is
Soo = > Oi.r 1 = for — 1 < r < 1
l: 1
76
CHAPTER 7. DIFFERENTIAL CALCULUS
7.2
where a\ is the first term of the series and r is the common ratio.
We see that there are some functions where the value of the function gets close to or approaches a
certain value.
Similarly, for the function:
x 2 + 4x  12
x + 6
The numerator of the function can be factorised as:
Qr + 6)(x2)
x + 6
Then we can cancel the x + 6 from numerator and denominator and we are left with:
However, we are only able to cancel the x + 6 term if x =£ —6. If x = —6, then the denominator
becomes and the function is not defined. This means that the domain of the function does not
include x = —6. But we can examine what happens to the values for y as x gets closer to —6. These
values are listed in Table 7.1 which shows that as x gets closer to —6, y gets close to 8.
Table 7.1 : Values for the function
(x + 6)(x2)
x + 6
as x gets close to 6.
X
_ (x + 6)(x2)
y x+6
9
li
8
10
7
9
6.5
8.5
6.4
8.4
6.3
8.3
6.2
8.2
6.1
8.1
6.09
8.09
6.08
8.08
6.01
8.01
5.9
7.9
5.8
7.8
5.7
7.7
5.6
7.6
5.5
7.5
5
7
4
6
3
5
The graph of this function is shown in Figure 7.1. The graph is a straight line with slope 1 and intercept
—2, but with a hole at x = —6.
7.2
CHAPTER 7. DIFFERENTIAL CALCULUS
Figure 7.1: Graph of y ■■
(ai + 6)(x2)
Limits
EMCBE
We can now introduce a new notation. For the function
(x + 6)(x2)
x + 6
we can write:
fan (3: + 6)( * 2) = 8.
This is read: the limit of < x + 6 >(^ 2 > as x tenc j s to _6 ; s _g_
CHAPTER 7. DIFFERENTIAL CALCULUS
7.2
Activity:
Limits
If /(a
1, determine:
/(o.i)
/(0.05)
/(0.04)
/(0.03)
/(0.02)
/(o.oi)
/(O.OO)
/(O.OI)
/(0.02)
/(0.03)
/(o.oi)
/(0.05)
/(o.i)
What do you notice about the value of f(x) as ie gets close to 0?
Example 1: Limits Notation
QUESTION
Summarise the following
the function
situation
by
using limit notation: As x gets close to 1, the value of
y = x + 2
gets close to 3.
SOLUTION
This is written as:
lim x + 2 = 3
X>1
in limit notation.
We can also have the situation where a function has a different value depending on whether x ap
proaches from the left or the right. An example of this is shown in Figure 7.2.
7.2
CHAPTER 7. DIFFERENTIAL CALCULUS
H h
H 1 1 fill
7 x
Figure 7.2: Graph of y = .
As x — y from the left, y =  approaches — oo. As x — > from the right, y =  approaches +oo. This
is written in limits notation as:
for x approaching from the left and
lim — = — oo
x^O X
lim — = oo
x+o+ x
for x approaching from the right. You can calculate the limit of many different functions using a set
method.
Method:
Limits: If you are required to calculate a limit like lim x ^„ then:
1 . Simplify the expression completely.
2. If it is possible, cancel all common terms.
3. Let a; approach a.
Example 2: Limits
QUESTION
Determine
lim 10
80
CHAPTER 7. DIFFERENTIAL CALCULUS 7.2
SOLUTION
Step 1 : Simplify the expression
There is nothing to simplify.
Step 2 : Cancel all common terms
There are no terms to cancel.
Step 3 : Let x — > 1 and write final answer
lim 10 = 10
xtl
Example 3: Limits
QUESTION
Determine
lim c
x^2
SOLUTION
Step 1 : Simplify the expression
There is nothing to simplify.
Step 2 : Cancel all common terms
There are no terms to cancel.
Step 3 : Let x — > 2 and write final answer
lim x = 2
x^2
81
7.2 CHAPTER 7. DIFFERENTIAL CALCULUS
Example 4: Limits
QUESTION
Determine
lim 
2  100
x10
SOLUTIO\
Step 1 :
Simplify the expression
The numerator can be factorised
x 2  100 (x + 10)(x
10)
x10
x10
Step 2 :
Cancel all common terms
(x — 10) can be cancelled from
the numerator and denominator.
(x + 10)(a;
x 10
10 ». I + io
Step 3 :
Let x — > 10 and write final answer
x 2 
lim
£►10 x 
H?»
Average Gradient and Gradient at a Point W emcbf
In Grade 10 you learnt about average gradients on a curve. The average gradient between any two
points on a curve is given by the gradient of the straight line that passes through both points. In Grade
1 1 you were introduced to the idea of a gradient at a single point on a curve. We saw that this was
the gradient of the tangent to the curve at the given point, but we did not learn how to determine the
gradient of the tangent.
Now let us consider the problem of trying to find the gradient of a tangent t to a curve with equation
y = /(x) at a given point P.
82
CHAPTER 7. DIFFERENTIAL CALCULUS
7.2
We know how to calculate the average gradient between two points on a curve, but we need two
points. The problem now is that we only have one point, namely P. To get around the problem we
first consider a secant to the curve that passes through point P and another point on the curve Q,
where Q is an arbitrary distance (h) from P, as shown in the figure. We can now find the average
gradient of the curve between points P and Q.
f(a + h)
f(a)
secant
If the xcoordinate of P is a + h, then the ^coordinate is /(a + h). Similarly, if the zcoordinate of Q
is a, then the ^coordinate is f(a). If we choose a + h as xi and a as xi, then:
2A = /0)
yi = f(a + h).
We can now calculate the average gradient as:
2/22/1 f(a + h)  f(a)
(a + h) — a
f(a + h)f(a)
h
(7.12)
(7.13)
Now imagine that Q moves along the curve toward P. The secant line approaches the tangent line as
its limiting position. This means that the average gradient of the secant approaches the gradient of the
tangent to the curve at P. In (??) we see that as point Q approaches point P, h gets closer to 0. When
h = 0, points P and Q are equal. We can now use our knowledge of limits to write this as:
gradient at P = lim
h>0
f(g + h)f(h)
h
(7.14)
and we say that the gradient at point P is the limit of the average gradient as Q approaches P along
the curve.
© See video: VMgye at www.everythingmaths.co.za
Example 5: Limits
QUESTION
For the function f(x) = 2x 2 — 5x, determine the gradient of the tangent to the curve at the
S:{
7.2 CHAPTER 7. DIFFERENTIAL CALCULUS
point x = 2.
SOLUTION
Step 1 : Calculating the gradient at a point
We know that the gradient at a point x is given by:
lim f{x + k)  f{x)
h»0 h
In our case x = 2. It is simpler to substitute x = 2 at the end of the calculation.
Step 2 : Write f(x + h) and simplify
f(x + h) = 2{x + h) 2 5(x + h)
= 2(x 2 + 2xh + h 2 )  5x  5h
= 2x 2 + 4xh + 2h 2 5x5h
Step 3 : Calculate limit
f(x + h)f(x) 2x 2 + Axh + 2h 2  5x  5h  (2x 2  5x) _
fc>o h h
,. 2x 2 + 4z/i + 2/i 2  5k  5/i  2x 2 + 5x
= lim
d»o h
&xh + 2/i 2  5/i
lim
hX)
lim
h»o /i
h(4x + 2h5)
lim Ax + 2h  5
= 4a;  5
Step 4 : Calculate gradient at x = 2
4x  5 = 4(2) 5 = 3
Step 5 : Write the final answer
The gradient of the tangent to the curve f(x) = 2x 2 — 5x at x = 2 is 3. This is
also the gradient of the curve at x = 2.
Example 6: Limits
QUESTION
For the function f(x) = 5x 2  4x + 1, determine the gradient of the tangent to curve at the
<sl
CHAPTER 7. DIFFERENTIAL CALCULUS
7.2
point x = a.
SOLUTION
Step 1 : Calculating the gradient at a point
We know that the gradient at a point x is given by:
lim
f(x + h)f(x)
In our case x = a. It is simpler to substitute x = a at the end of the calculation.
Step 2 : Write f(x + ft) and simplify
f(x + h) = 5(x + hf  A(x + ft) + 1
= 5(x 2 + 2xft + ft 2 ) 4x4ft + l
= 5x 2 + lOxh + 5ft 2  4x  4ft + 1
Step 3 : Calculate limit
/(x + ft)/(x)
lim
7it0
ft
5x 2 + lOxft + 5ft 2  4x  4ft + 1  (5x 2  4x + 1)
ft
5x 2 + lOxft + 5ft 2  4x  4ft + 1  5x 2 + 4x  1
lim
h to
lim
ft to
lim
ftto ft
lim lOx + 5ft  4
10x4
10xft + 5ft 2 4ft
ft(10x + 5ft4)
Step 4 : Calculate gradient at x = a
10x  4 = 10a  4
Step 5 : Write the final answer
The gradient of the tangent to the curve /(x) = 5x 2 — Ax + 1 at x = 1 is 10a — 4.
Exercise 71
Determine the following
1.
2.
lim
x 2 9
at3 X + 3
lim
x + 3
xt3 X 2 + 3x
85
7.3
CHAPTER 7. DIFFERENTIAL CALCULUS
3.
lim
3z  4x
lim
2 3x
x 2 x 12
+ 4 x  4
lim 3z + —
x^2 3x
Aj More practice (►) video solutions Cf) or help at www.everythingmaths
(l.)01fh (2.)01fi (3.) 01 fj (4.) 01 fk (5.) 01fm
7.3 Differentiation from First
Principles
The tangent problem has given rise to the branch of calculus called differential calculus and the
equation:
Um f(x + h) f(x)
fcMJ h
defines the derivative of the function f(x). Using (7.15) to calculate the derivative is called finding
the derivative from first principles.
Tip
Though we choose to
use a fractional form of
representation, jg is a
limit and is not a frac
tion, i.e. §^ does not
' dx
mean dy f dx. =^
" dx
means y differentiated
with respect to x. Thus,
^ means p differenti
ated with respect to x.
ator", applied to some
function of x.
DEFINITION: Derivative
The derivative of a function f(x) is written as f'(x) and is defined by:
f{x) = lim lk+K >/w
(7.15)
There are a few different notations used to refer to derivatives. If we use the traditional notation
y = f{x) to indicate that the dependent variable is y and the independent variable is x, then some
common alternative notations for the derivative are as follows:
/'(*)
.'/
dy _ df_
dx dx
dx
f{x) = Df(x) = D x f(x)
The symbols D and 4 are called differential operators because they indicate the operation of differ
entiation, which is the process of calculating a derivative. It is very important that you learn to identify
these different ways of denoting the derivative, and that you are consistent in your usage of them when
answering questions.
See video: VMgyp at www.everythingmaths.co.za
Mi
CHAPTER 7. DIFFERENTIAL CALCULUS 7.3
Example 7: Derivatives  First Principles
QUESTION
Calculate the derivative of g(x) = x — 1 from first principles.
SOLUTION
Step I : Calculate the gradient at a point
We know that the gradient at a point x is given by:
// x , 9(x + h) g(x)
q (x) = urn — 7 ~^
Step 2 : Write g(x + ft) and simplify
g(x + ft) = x + ft — 1
Step 3 : Calculate limit
1 1 x ,. g(x + h)  g(x)
q (x) = lim — i ^
x + h 
1
(xl)
h¥0
ft
x + h
 1 
x + 1
hK3
ft
lim —
h^O h
lim 1
1
Step 4 : Wrife f/ie final answer
The derivative g'(a;) of g(x) = x — 1 is 1.
Exercise 72
1 . Given g(x) = — x 2
, , , . o(a: + ft) — o(a;)
(a) determine ^ / — ! ^— L
h
(b) hence, determine
Um g(x + h)g(x)
h»0 ft
(c) explain the meaning of your answer in (b).
2. Find the derivative of f(x) = —2x 2 + 3x using first principles.
87
7.4
CHAPTER 7. DIFFERENTIAL CALCULUS
3. Determine the derivative of f(x)
x2
using first principles.
4. Determine /'(3) from first principles if f(x) = —5a; 2 .
5. If h(x) = 4a: 2 — Ax, determine h'(x) using first principles.
CjX*y More practice f ►) video solutions f? J or help at www.everythingmaths.co.za
(1.)01fn (2.)01fp (3.) 01fq (4.) 01fr (5.) 01fs
7.4 Rules of Differentiation
Calculating the derivative of a function from first principles is very long, and it is easy to make mistakes.
Fortunately, there are rules which make calculating the derivative simple.
Activity:
Rules of Differentiation
From first principles, determine the derivatives of the following:
1. f(x) = b
2. f(x) = x
3. f(x) = x 2
4. f(x) = x 3
5. f(x) = l/x
You should have found the following:
/(*)
/'(*)
b
X
1
x 1
2x
x :i
3x 2
l/x = x~ L
—x
If we examine these results we see that there is a pattern, which can be summarised by:
dx
(X ) = nx
(7.16)
There are two other rules which make differentiation simpler. For any two functions f(x) and g(x):
d
dx
lf(x)±g(x)} = f'(x)±g'(x)
This means that we differentiate each term separately.
The final rule applies to a function f(x) that is multiplied by a constant k.
^ x [k.f(x)} = kf'(x)
See video: VMhgb at www.everythingmaths.co.za
(7.17)
(7.18)
,ss
CHAPTER 7. DIFFERENTIAL CALCULUS
7.4
Example 8: Rules of Differentiation
QUESTION
Determine the derivative of x — 1 using the rules of diffei
entiation.
SOLUTION
Step 1
Identify the rules that will be needed
We will apply two rules of differentiation:
0* t n \ n — 1
— — (x I = nx
dx
and
i\m a w = i\m
■5W*M
Step 2
Determine the derivative
In our case f(x) = x and g(x) = 1.
fix) = 1
and
g'(x) =
Step 3
Write the final answer
The derivative of x — 1 is 1 which is the same result as was obtained earlier, from
first principles.
Given two functions, f(x) and g(x) and constant b, n and k, we know that:
^6 =
£(x")=nxi
dx V J I dx
dx w + 9) dx + dx
89
7.5
CHAPTER 7. DIFFERENTIAL CALCULUS
Exercise 73
1. Find/'(x)if/(x) = ^
5x + 6
2
2. Find f'(y) if f(y) = rfj.
3. Kndf'(z)\ff(z) = (zl)(z + l).
4. Determine f if y = ** + 2 ^ ~ 3 .
5. Determine the derivative of y = ya? \
3a;
l\n More practice (►) video solutions (9j or help at www.everythingmaths
(1.) 01ft (2.)01fu (3.)01fv (4.)01fw (5.)01fx
7.5 Applying Differentiation to
Draw Graphs
EMCB]
Thus far we have learnt about how to differentiate various functions, but I am sure that you are be
ginning to ask, What is the point of learning about derivatives? Well, we know one important fact
about a derivative: it is a gradient. So, any problems involving the calculations of gradients or rates of
change can use derivatives. One simple application is to draw graphs of functions by firstly determine
the gradients of straight lines and secondly to determine the turning points of the graph.
Finding Equations of Tangents to Curves
EMCBK
In Section 7.2.4 we saw that finding the gradient of a tangent to a curve is the same as finding the
gradient (or slope) of the same curve at the point of the tangent. We also saw that the gradient of a
function at a point is just its derivative.
Since we have the gradient of the tangent and the point on the curve through which the tangent passes,
we can find the equation of the tangent.
Oil
CHAPTER 7. DIFFERENTIAL CALCULUS
7.5
Example 9: Finding the Equation of a Tangent to a Curve
QUESTION
Find the equation of the tangent to the curve y = x 2 at the point (1; 1) and draw both
functions.
SOLUTION
Step 1 : Determine what is required
We are required to determine the equation of the tangent to the curve defined
by y = x 2 at the point (1; 1). The tangent is a straight line and we can find the
equation by using derivatives to find the gradient of the straight line. Then we
will have the gradient and one point on the line, so we can find the equation
using:
y — j/i = m(x — xi)
from Grade 1 1 Coordinate Geometry.
Step 2 : Differentiate the function
Using our rules of differentiation we get:
y = 2x
Step 3 : Find the gradient at the point (1; 1)
In order to determine the gradient at the point (1: 1), we substitute the xvalue
into the equation for the derivative. So, y' at x = 1 is:
m = 2(1) = 2
Step 4 : Find the equation of the tangent
lVi
=
m{x — Xi)
01
=
(2)(xl)
y
=
2x  2 + 1
y
=
2x 1
Step 5 : Write the final answer
The equation of the tangent to the curve defined by y = x 2 at the point (1; 1) is
y = 2x  1.
Step 6 : Sketch both functions
91
7.5
CHAPTER 7. DIFFERENTIAL CALCULUS
\ 1 1 h
4 3 2 1
y = x
y = 2x  1
Curve Sketching
EMCBL
Tip
If X
= a is
a turning
point of f(x),
then:
/'(a) =
=
This
means
that the
derivative is
at a
turnir
g point.
Differentiation can be used to sketch the graphs of functions, by helping determine the turning points.
We know that if a graph is increasing on an interval and reaches a turning point, then the graph will
start decreasing after the turning point. The turning point is also known as a stationary point because
the gradient at a turning point is 0. We can then use this information to calculate turning points, by
calculating the points at which the derivative of a function is 0.
Take the graph of y = x 2 as an example. We know that the graph of this function has a turning point
at (0,0), but we can use the derivative of the function:
y = 2x
and set it equal to to find the zvalue for which the graph has a turning point.
2x =
x =
We then substitute this into the equation of the graph (i.e. y = x 2 ) to determine the ^coordinate of
the turning point:
/(0) = (0) 2 =
This corresponds to the point that we have previously calculated.
92
CHAPTER 7. DIFFERENTIAL CALCULUS 7.5
Example 10: Calculation of Turning Points
QUESTION
Calculate the turning points of the graph of the function
f(x) = 2x 3  9x 2 + 12x  15
SOLUTION
Step I : Determine the derivative of f(x)
Using the rules of differentiation we get:
f'(x) = fix 2  18x+ 12
Step 2 : Set f'(x) = and calculate xcoordinate of turning point
6x  18x + 12
=
x 2  3x + 2
=
(x2)(xl)
=
Therefore, the turning points are at x = 2 and x = 1.
Step 3 : Substitute xcoordinate of turning point into f(x) to determine ycoordinates
/(2) = 2(2) 3 9(2) 2 + 12(2) 15
= 1636 + 2415
= 11.
/(I) = 2(1) J 9(1) 2 + 12(1)15
= 29 + 1215
= 10
Step 4 : Write final answer
The turning points of the graph of /(x) = 2x 3 — 9x 2 + 12x — 15 are (2; —11)
and (1;10).
We are now ready to sketch graphs of functions.
Method:
Sketching Graphs: Suppose we are given that f(x) = ax 3 + bx 2 + ex + d, then there are five steps to
be followed to sketch the graph of the function:
1 . Determine the value of the ?/intercept by substituting x = into f(x)
93
7.5 CHAPTER 7. DIFFERENTIAL CALCULUS
2. Determine the xintercepts by factorising ax 3 + bx 2 + ex + d = and solving for x. First try
to eliminate constant common factors, and to group like terms together so that the expression is
expressed as economically as possible. Use the factor theorem if necessary.
3. Find the turning points of the function by working out the derivative ^ and setting it to zero,
and solving for x.
4. Determine the j/coordinates of the turning points by substituting the x values obtained in the
previous step, into the expression for f(x).
5. Use the information you're given to plot the points and get a rough idea of the gradients between
points. Then fill in the missing parts of the function in a smooth, continuous curve.
Example 11: Sketching Graphs
QUESTION
Draw the graph of g(x) = x 2 — x + 2
SOLUTION
Step 7 : Determine the shape of the graph
The leading coefficient of x is > therefore the graph is a parabola with a
minimum.
5tep 2 : Determine the yintercept
The ^intercept is obtained by setting x = 0.
9 (0) = (0) 2 0 + 2 = 2
The turning point is at (0: 2).
Step 3 : Determine the xintercepts
The xintercepts are found by setting g{x) = 0.
g(x) = x — x + 2
= x 2  x + 2
Using the quadratic formula and looking at b 2 — iac we can see that this would
be negative and so this function does not have real roots. Therefore, the graph of
g(x) does not have any xintercepts.
Step 4 : Find the turning points of the function
Work out the derivative ^ and set it to zero to for the x coordinate of the turning
point.
^=2xl
ax
dx
2x 1
2x = 1
1
X = 2
94
CHAPTER 7. DIFFERENTIAL CALCULUS
7.5
Step 5 : Determine the ycoordinates of the turning points by substituting the x values
obtained in the previous step, into the expression for f(x).
y coordinate of turning point is given by calculating g{\).
V
1 1
4 ~ 2
7
4
The turning point is at (; J)
Step 6 : Draw a neat sketch
3 2 1
I 1 1 h» *
12 3 4
Example 12: Sketching Graphs
QUESTION
Sketch the graph of g(x) = x 3 + 6x 2  9x + 4.
SOLUTION
Step 1 : Determine the shape of the graph
The leading coefficient of x is < therefore the graph is a parabola with a
maximum.
95
7.5 CHAPTER 7. DIFFERENTIAL CALCULUS
Step 2 : Determine the yintercept
We find the yintercepts by finding the value for 9(0).
g(x) = — x + 6x — 9x + 4
y int = g(0) = (0) 3 +6(0) 2 9(0)+4
= 4
Step 3 : Determine the xintercepts
We find the xintercepts by finding the points for which the function g(x) = 0.
g(x) = —x + 6x — 9x + 4
Use the factor theorem to confirm that (x — 1) is a factor. If g(l) = 0, then
(x — 1) is a factor.
g(x) = —x + 6x — 9x + 4
9(1) = (l) 3 +6(l) 2 9(l) + 4
= 1 + 69 + 4
=
Therefore, (x — 1) is a factor.
If we divide g(x) by (x — 1) we are left with:
—x + 5x — 4
This has factors
(x4)(x 1)
Therefore:
9 (x) = (xl)(xl)(x4)
The xintercepts are: x int = 1; 4
Step 4 : Calculate the turning points
Find the turning points by setting g'(x) = 0.
If we use the rules of differentiation we get
g'(x) = 3x 2 + 12x9
g'(x) =
3x 2 + 12x  9 =
x 2  4x + 3 =
(x3)(xl) =
clearpage The xcoordinates of the turning points are: x = 1 and x = 3.
The ^/coordinates of the turning points are calculated as:
9{x)
=
—x + 6x — 9x
+ 4
9(1)
=
(1) 3 +6(1) 2 
9(1) + 4
=
1+69+4
=
96
CHAPTER 7. DIFFERENTIAL CALCULUS
7.5
g(x) = — x + 6x — 9x + 4
3(3) = (3) 3 +6(3) 2 9(3) + 4
= 27 + 5427 + 4
= 4
Therefore the turning points are: (1; 0) and (3; 4).
Step 5 : Draw a neat sketch
Exercise 74
1 . Given f(x) = x 3 + x 2  5x + 3:
(a) Show that (x — 1) is a factor of f(x) and hence factorise f(x) fully.
(b) Find the coordinates of the intercepts with the axes and the turning points and sketch the
graph
2. Sketch the graph of f(x) = x :i — Ax 1 — llx + 30 showing all the relative turning points and
intercepts with the axes.
3. (a) Sketch the graph of f(x) = x 3 — 9x 2 + 24x — 20, showing all intercepts with the axes and
turning points.
(b) Find the equation of the tangent to f(x) at x = 4.
97
7.5
CHAPTER 7. DIFFERENTIAL CALCULUS
l\n More practice Cr) video solutions f'fj or help at www.everythingmaths.co.za
(1.)01fy (2.)01fz (3.)01gO
Local Minimum, Local Maximum and Point
of Inflection
EMCBM
If the derivative (J 2 ) is zero at a point, the gradient of the tangent at that point is zero. It means that a
turning point occurs as seen in the previous example.
From the drawing the point (1; 0) represents a local minimum and the point (3; 4) the local maximum.
A graph has a horizontal point of inflexion where the derivative is zero but the sign of the gradient
does not change. That means the graph will continue to increase or decrease after the stationary point.
MS
CHAPTER 7. DIFFERENTIAL CALCULUS
7.6
From this drawing, the point (3; 1) is a horizontal point of inflexion, because the sign of the derivative
does not change from positive to negative.
7.6 Using Differential Calculus to
Solve Problems
We have seen that differential calculus can be used to determine the stationary points of functions, in
order to sketch their graphs. However, determining stationary points also lends itself to the solution of
problems that require some variable to be optimised.
For example, if fuel used by a car is defined by:
^v 2 6v + 245
oU
(7.19)
where v is the travelling speed, what is the most economical speed (that means the speed that uses the
least fuel)?
If we draw the graph of this function we find that the graph has a minimum. The speed at the minimum
would then give the most economical speed.
60
§ 50
c
O
E. 40
E
 30
o
u
1 20
10
o + — I 1 1 1 1 1 1 1 1 1 1 1 1 \—
10 20 30 40 50 60 70 80 90 100 110 120 130 140
speed (km/h)
99
7.6 CHAPTER 7. DIFFERENTIAL CALCULUS
We have seen that the coordinates of the turning point can be calculated by differentiating the function
and finding the xcoordinate (speed in the case of the example) for which the derivative is 0.
Differentiating (7.19), we get:
f'(v) = v6
If we set f'(v) = we can calculate the speed that corresponds to the turning point.
° = l V  6
6 x 40
V = 3~
= 80
This means that the most economical speed is 80 km/h.
See video: VMhgi at www.everythingmaths.co.za
Example 13: Optimisation Problems
QUESTION
The sum of two positive numbers is 10. One of the numbers is multiplied by the square of the
other. If each number is greater than 0, find the numbers that make this product a maximum.
SOLUTION
Step 1
Examine the problem and formulate the equations that are required
Let the two numbers be a and b. Then we have:
a + b = 10
(7.20)
We are required to minimise the product of a and b. Call the proc
Then:
uct P.
P = a.b
(7.21)
We can solve for b from (7.20) to get:
b = 10  a
(7.22)
Substitute this into (7.21) to write P in terms of a only.
P = a (10  a) = 10a  a 2
(7.23)
Step 2
Differentiate
The derivative of (7.23) is:
P'(a) = 10  2a
Step 3
Find the stationary point
Set P'{a) = to find the value of a which makes P a maximum.
100
CHAPTER 7. DIFFERENTIAL CALCULUS
7.6
P'(a)
=
10
2a
=
10
2a
2a
=
10
a
=
10
2
a
=
5
Substitute into (7.26) to solve for the width.
b = 10  a
= 105
= 5
Step 4 : Write the final answer
The product is maximised when a and b are both equal to 5.
Example 14: Optimisation Problems
QUESTION
Michael wants to start a vegetable garden, which he decides to fence off in the shape of a
rectangle from the rest of the garden. Michael has only 160 m of fencing, so he decides to use
a wall as one border of the vegetable garden. Calculate the width and length of the garden
that corresponds to largest possible area that Michael can fence off.
width, w
SOLUTION
Step I : Examine the problem and formulate the equations that are required
The important pieces of information given are related to the area and modified
perimeter of the garden. We know that the area of the garden is:
.1
(7.24)
We are also told that the fence covers only 3 sides and the three sides should add
up to 160 m. This can be written as:
160 m = w + 1 + 1
(7.25)
101
7.6 CHAPTER 7. DIFFERENTIAL CALCULUS
However, we can use (7.25) to write w in terms of /:
w = 160m 21 (7.26)
Substitute (7.26) into (7.24) to get:
A = (160 m  21)1 = 160 m  21 2 (7.27)
Step 2 : Differentiate
Since we are interested in maximising the area, we differentiate (7.27) to get:
A' (I) = 160 m 41
Step 3 : Find the stationary point
To find the stationary point, we set A'(l) = and solve for the value of I that
maximises the area.
A' (I) = 160 m 41
= 160m 41
:Al = 160m
160 m
4
I = 40 m
Substitute into (7.26) to solve for the width.
w = 160 m 21
= 160 m 2(40 m)
= 160 m 80 m
= 80 m
Step 4 : Write the final answer
A width of 80 m and a length of 40 m will yield the maximal area fenced off.
Exercise 75
1 . The sum of two positive numbers is 20. One of the numbers is multiplied by the square of the
other. Find the numbers that make this product a maximum.
2. A wooden block is made as shown in the diagram. The ends are rightangled triangles having
sides 3x, Ax and 5x. The length of the block is y. The total surface area of the block is 3600 cm 2 .
102
CHAPTER 7. DIFFERENTIAL CALCULUS
7.6
(a) Show that y
1200cm 2  4x 2
ox
(b) Find the value of x for which the block will have a maximum volume. (Volume = area of
base x height.)
3. The diagram shows the plan for a veranda which is to be built on the corner of a cottage. A
railing ABCDE is to be constructed around the four edges of the veranda.
c
y
D
J
L
verandah
X
1
F
_C
B
r
J
A
E
cottage
If AB = DE = x and BC = CD = y, and the length of the railing must be 30 m, find the
values of x and y for which the verandah will have a maximum area.
More practice f ►) video solutions CfJ or help at www.everythingmaths.co.za
(1.)01g4 (2.)01g5 (3.) 01;
Rate of Change Problems
EMCBO
Two concepts were discussed in this chapter: Average rate of change = f(  b l _^ (a) and Instantaneous
rate of change = lim^o
f(x + h).f(x)
When we mention rate of change, the latter is implied. Instan
11).',
7.6 CHAPTER 7. DIFFERENTIAL CALCULUS
taneous rate of change is the derivative. When average rate of change is required, it will be specifically
referred to as average rate of change.
Velocity is one of the most common forms of rate of change. Again, average velocity = average rate
of change and instantaneous velocity = instantaneous rate of change = derivative. Velocity refers to
the increase of distance(s) for a corresponding increase in time (t). The notation commonly used for
this is:
v(t) = J = ,'(*)
where s'(t) is the position function. Acceleration is the change in velocity for a corresponding increase
in time. Therefore, acceleration is the derivative of velocity
a(t) = v'(t)
This implies that acceleration is the second derivative of the distance(s).
Example 15: Rate of Change
QUESTION
The height (in metres) of a golf ball that is hit into the air after t seconds, is given by h(t)
20*  5t 2 . Determine
1 . the average velocity of the ball during the first two seconds
2. the velocity of the ball after 1,5 s
3. the time at which the velocity is zero
4. the velocity at which the ball hits the ground
5. the acceleration of the ball
SOLUTION
Step 1 : Average velocity
= 10ms
Step 2 : Instantaneous Velocity
v(t)
h{2)  /t(0)
20
[20(2)  5(2) 2 ]  [20(0)  5(0) 2
2
4020
2
l
dh
dt
20  10*
104
CHAPTER 7. DIFFERENTIAL CALCULUS
7.6
Velocity after 1,5 s:
v(l,5) = 2010(1,5)
r 1
= 5m ■ s
Step 3 : Zero velocity
v(t)
=
20
 104
=
Wt
=
20
t
=
2
Therefore the velocity is zero after 2 s
Step 4 : Ground velocity
The ball hits the ground when hit) =
20*  5i 2 =
5t(4  1) =
t = or t = 4
The ball hits the ground after 4s. The velocity after 4s will be:
1,(4) = h'{A)
= 20  10(4)
= 20 ms 1
The ball hits the ground at a speed of 20 m ■ s _1 . Notice that the sign of the
velocity is negative which means that the ball is moving downward (the reverse
of upward, which is when the velocity is positive).
Step 5 : Acceleration
a = v (t)
= —10 m ■ s~
Just because gravity is constant does not mean we should necessarily think of
acceleration as a constant. We should still consider it a function.
Chapter 7
End of Chapter Exercises
1 . Determine f'(x) from first principles if:
105
7.6 CHAPTER 7. DIFFERENTIAL CALCULUS
3. Determine ^f if:
dy
2. Given: f(x)
Dete
(a)
(b)
f(x) = x 
 (ix
f(x) = 2x 
2
 X
3.x,
find
f'(x) using first print
ziples
y = (2x) 2 
1
3.x
2V5
5
\/x
2
4. Given: f(x) = x 3  3x 2 + 4
(a) Calculate /(— 1), and hence solve the equation /(x) =
(b) Determine f'(x)
(c) Sketch the graph of / neatly and clearly, showing the coordinates of the turning
points as well as the intercepts on both axes.
(d) Determine the coordinates of the points on the graph of / where the gradient is
9.
5. Given: /(x) = 2x 3 — 5x 2 — 4x + 3. The xintercepts of / are: (— 1;0) (;0) and
(3;0).
(a) Determine the coordinates of the turning points of /.
(b) Draw a neat sketch graph of /. Clearly indicate the coordinates of the intercepts
with the axes, as well as the coordinates of the turning points.
(c) For which values of k will the equation /(x) = k , have exactly two real roots?
(d) Determine the equation of the tangent to the graph of f(x) = 2x 3 — 5x 2 — 4x + 3
at the point where x = 1.
6. (a) Sketch the graph of /(x) = x 3 — 9x 2 + 24x — 20, showing all intercepts with the
axes and turning points,
(b) Find the equation of the tangent to /(x) at x = 4.
7. Calculate:
1x 3
lim
xn 1 — x
8. Given:
fix) = 2x — x
(a) Use the definition of the derivative to calculate f'(x).
(b) Hence, calculate the coordinates of the point at which the gradient of the tan
gent to the graph of / is 7.
9. If xy — 5 = vV, determine j^
10. Given: g(x) = (x~ 2 +x 2 ) 2 . Calculate g' (2).
11. Given: /(x) = 2x  3
(a) Find: / _1 (x)
(b) Solve: / _1 (x) = 3/'(x)
12. Find f'(x) for each of the following:
(a) fix) = ^ + 10
(b) f(x) = (2x»5)(3x + 2)
x 2
1 3. Determine the minimum value of the sum of a positive number and its reciprocal.
14. If the displacement s (in metres) of a particle at time t (in seconds) is governed by the
equation s = \t 3 — 2t, find its acceleration after 2 seconds. (Acceleration is the rate
of change of velocity, and velocity is the rate of change of displacement.)
106
CHAPTER 7. DIFFERENTIAL CALCULUS 7.6
1 5. (a) After doing some research, a transport company has determined that the rate at
which petrol is consumed by one of its large carriers, travelling at an average
speed of x km per hour, is given by:
55 a 1
P(x) ■■
2x 200 km 2 ^ 1 !! 1
i. Assume that the petrol costs R4,00 per litre and the driver earns R18,00 per
hour (travelling time). Now deduce that the total cost, C, in Rands, for a
2 000 km trip is given by:
_, , 256000 kmh _1 R ,„
C(x) = h 40x
x
ii. Hence determine the average speed to be maintained to effect a minimum
cost for a 2 000 km trip.
(b) During an experiment the temperature T (in degrees Celsius), varies with time t
(in hours), according to the formula:
T{t) = 30 + 4t i* 2 te[l;10]
i. Determine an expression for the rate of change of temperature with time,
ii. During which time interval was the temperature dropping?
16. The depth, d, of water in a kettle t minutes after it starts to boil, is given by d =
86 — it — jt\ where d is measured in millimetres.
(a) How many millimetres of water are there in the kettle just before it starts to boil?
(b) As the water boils, the level in the kettle drops. Find the rate at which the water
level is decreasing when t = 2 minutes.
(c) How many minutes after the kettle starts boiling will the water level be dropping
at a rate of 12 mm/minute?
(ft" 1 ) More practice (►) video solutions CfJ or help at www.everythingmaths.co.za
(1.)01g7 (2.)01g8 (3.)01g9 (4.)01ga (5.) 01gb (6.) Olgc
(7.)01gd (8.)01ge (9.)01gf (10.) 01gg (11.)01gh (12.) 01 gi
(13.)01gj (14.)01gk (15.)01gm (16.)021h
107
Linear Programming
8. 1 Introduction
In Grade 11 you were introduced to linear programming and solved problems by looking at points
on the edges of the feasible region. In Grade 12 you will look at how to solve linear programming
problems in a more general manner.
© See introductory video: VMhgw at www.everythingmaths.co.za
8.2 Terminology
Feasible Region and Points
EMCBR
Tip
The constraints are used
to create bounds of the
solution.
Tip
ax + by = c
o If b ^ 0, feasible
points must lie on the
o If b = 0, feasible
points must lie on the
line x = c/a
ax + by < c
o If b jL 0, feasible
points must lie on
or below the line
o If b = 0, feasible
points must lie on or
to the left of the line
x = c/a.
Constraints mean that we cannot just take any x and y when looking for the x and y that optimise our
objective function. If we think of the variables x and y as a point (x; y) in the xj/plane then we call
the set of all points in the xj/plane that satisfy our constraints the feasible region. Any point in the
feasible region is called a feasible point.
For example, the constraints
x >
y>0
mean that every (x,y) we can consider must lie in the first quadrant of the xy plane. The constraint
x>y
means that every (x,y) must lie on or below the line y = x and the constraint
x < 20
means that x must lie on or to the left of the line x = 20.
We can use these constraints to draw the feasible region as shown by the shaded region in Figure 8.1 .
When a constraint is linear, it means that it requires that any feasible point (x,y) lies on one side of
or on a line. Interpreting constraints as graphs in the xy plane is very important since it allows us to
construct the feasible region such as in Figure 8.1.
108
CHAPTER 8. LINEAR PROGRAMMING
8.3
5 10 15 20
Figure 8.1: The feasible region corresponding to the constraints x >0, y >0, x > y and x < 20.
8.3 Linear Programming and the
Feasible Region
If the objective function and all of the constraints are linear then we call the problem of optimising
the objective function subject to these constraints a linear program. All optimisation problems we will
look at will be linear programs.
The major consequence of the constraints being linear is that the feasible region is always a polygon.
This is evident since the constraints that define the feasible region all contribute a line segment to its
boundary (see Figure 8.1). It is also always true that the feasible region is a convex polygon.
The objective function being linear means that the feasible point(s) that gives the solution of a linear
program always lies on one of the vertices of the feasible region. This is very important since, as we
will soon see, it gives us a way of solving linear programs.
We will now see why the solutions of a linear program always lie on the boundary of the feasible
region. Firstly, note that if we think of f(x,y) as lying on the z axis, then the function f(x,y) = ax + by
(where a and b are real numbers) is the definition of a plane. If we solve for y in the equation defining
the objective function then
f(x,y)
ax + by
f(x,y)
(8.1)
What this means is that if we find all the points where f(x,y) = c for any real number c (i.e. f(x,y)
is constant with a value of c), then we have the equation of a line. This line we call a level line of the
objective function.
Consider again the feasible region described in Figure 8.1. Let's say that we have the objective function
f(x,y) = x — 2y with this feasible region. If we consider Equation 8.3 corresponding to
then we get the level line
f(x,y) = 20
y = x + 10
y 2
10!)
8.3
CHAPTER 8. LINEAR PROGRAMMING
which has been drawn in Figure 8.2. Level lines corresponding to
x
f(x,y) = 10
or
y=2 + 5
/(*,!/) =
or
X
V= 2
f(x,y) = 10
or
X K
y = x — 5
y 2
f(x,y) = 20
or
»=fio
have also been drawn in. It is very important to realise that these are not the only level lines; in fact,
there are infinitely many of them and they are all parallel to each other. Remember that if we look at
any one level line f(x,y) has the same value for every point (x,y) that lies on that line. Also, f(x,y)
will always have different values on different level lines.
10 AS
Figure 8.2: The feasible region corresponding to the constraints x > 0, y > 0, x > y and x < 20 with
objective function f(x,y) = x — 2y, The dashed lines represent various level lines of f(x,y).
If a ruler is placed on the level line corresponding to f(x,y) = —20 in Figure 8.2 and moved down the
page parallel to this line then it is clear that the ruler will be moving over level lines which correspond
to larger values of f(x,y). So if we wanted to maximise f(x,y) then we simply move the ruler down the
page until we reach the bottommost point in the feasible region. This point will then be the feasible
point that maximises f(x,y). Similarly, if we wanted to minimise f(x,y) then the topmost feasible
point will give the minimum value of f(x,y).
Since our feasible region is a polygon, these points will always lie on vertices in the feasible region.
The fact that the value of our objective function along the line of the ruler increases as we move it
down and decreases as we move it up depends on this particular example. Some other examples
might have that the function increases as we move the ruler up and decreases as we move it down.
It is a general property, though, of linear objective functions that they will consistently increase or
decrease as we move the ruler up or down. Knowing which direction to move the ruler in order to
maximise/minimise f(x,y) = ax + by is as simple as looking at the sign of 6 (i.e. "is b negative, positive
or zero?"). If b is positive, then f(x,y) increases as we move the ruler up and f(x,y) decreases as we
move the ruler down. The opposite happens for the case when b is negative: f(x,y) decreases as we
move the ruler up and f(x,y) increases as we move the ruler down. If b = 0, we need to look at the
sign of a.
If a is positive then f(x,y) increases as we move the ruler to the right and decreases if we move the
ruler to the left. Once again, the opposite happens for a negative. If we look again at the objective
function mentioned earlier,
f(x,y) = x  2y
with a = 1 and b = —2, then we should find that f(x,y) increases as we move the ruler down the
page since b = — 2 < 0. This is exactly what happened in Figure 8.2.
The main points about linear programming we have encountered so far are
110
CHAPTER 8. LINEAR PROGRAMMING 8.3
• The feasible region is always a polygon.
• Solutions occur at vertices of the feasible region.
• Moving a ruler parallel to the level lines of the objective function up/down to the top/bottom of
the feasible region shows us which of the vertices is the solution.
• The direction in which to move the ruler is determined by the sign of b and also possibly by the
sign of a.
These points are sufficient to determine a method for solving any linear program.
Method: Linear Programming
If we wish to maximise the objective function f(x,y) then:
1 . Find the gradient of the level lines of f(x,y) (this is always going to be —  as we saw in Equa
tion ??)
2. Place your ruler on the xy plane, making a line with gradient —  (i.e. b units on the xaxis and
—a units on the yaxis)
3. The solution of the linear program is given by appropriately moving the ruler. Firstly we need to
check whether b is negative, positive or zero.
(a) If b > 0, move the ruler up the page, keeping the ruler parallel to the level lines all the time,
until it touches the "highest" point in the feasible region. This point is then the solution.
(b) If b < 0, move the ruler in the opposite direction to get the solution at the "lowest" point in
the feasible region.
(c) If b = 0, check the sign of a
i. If a < move the ruler to the "leftmost" feasible point. This point is then the solution,
ii. If a > move the ruler to the "rightmost" feasible point. This point is then the solution.
Example 1: Prizes!
QUESTION
As part of their opening specials, a furniture store promised to give away at least 40 prizes
with a total value of at least R2 000. The prizes are kettles and toasters.
1 . If the company decides that there will be at least 10 of each prize, write down two more
inequalities from these constraints.
2. If the cost of manufacturing a kettle is R60 and a toaster is R50, write down an objective
function C which can be used to determine the cost to the company of both kettles and
toasters.
3. Sketch the graph of the feasibility region that can be used to determine all the possible
combinations of kettles and toasters that honour the promises of the company.
4. How many of each prize will represent the cheapest option for the company?
5. How much will this combination of kettles and toasters cost?
SOLUTION
111
8.3
CHAPTER 8. LINEAR PROGRAMMING
Step 1 : Identify the decision variables
Let the number of kettles be x and the number of toasters be y and write down
two constraints apart from x > and y > that must be adhered to.
Step 2 : Write constraint equations
Since there will be at least 10 of each prize we can write:
x > 10
and
y> 10
Also, the store promised to give away at least 40 prizes in total. Therefore:
x + y > 40
Step 3 : Write the objective function
The cost of manufacturing a kettle is R60 and a toaster is R50. Therefore the cost
the total cost C is:
C = 60x + 50j/
Step 4 : Sketch the graph of the feasible region
i
90 
80 
70 
60 
50 
40 
30 
B
20 
\ A
\
1
1
20
i
30
40
i
50
60
70
80
90
100
Step 5 : Determine vertices of feasible region
From the graph, the coordinates of vertex A are (30; 10) and the coordinates of
vertex Bare (10; 30).
Step 6 : Draw in the search line
The search line is the gradient of the objective function. That is, if the equation
C = dOx + 50y is now written in the standard form y = . . ., then the gradient is:
6
m = ,
which is shown with the broken line on the graph.
112
CHAPTER 8. LINEAR PROGRAMMING
8.3
10 20 30 40 50 60 70 80 90 100
Step 7 : Calculate cost at each vertex
At vertex A, the cost is:
C = 60x + 50/
= 60(30) + 50(10)
= 1 800 + 500
= R2 300
At vertex B, the cost is:
C = 60x fc + 50»/t
= 60(10) + 50(30)
= 600 + 1500
= R2 100
Step 8 : Write the final answer
The cheapest combination of prizes is 10 kettles and 30 toasters, costing the
company R2 100.
Example 2: Search Line Method
QUESTION
As a production planner at a factory manufacturing lawn cutters your job will be to advise
the management on how many of each model should be produced per week in order to
maximise the profit on the local production. The factory is producing two types of lawn
cutters: Quadrant and Pentagon. Two of the production processes that the lawn cutters must
113
8.3
CHAPTER 8. LINEAR PROGRAMMING
go through are: bodywork and engine work.
• The factory cannot operate for less than 360 hours on engine work for the lawn cutters.
• The factory has a maximum capacity of 480 hours for bodywork for the lawn cutters.
• Half an hour of engine work and half an hour of bodywork is required to produce one
Quadrant.
• The ratio of Pentagon lawn cutters to Quadrant lawn cutters produced per week must be
at least 3 : 2.
• A minimum of 200 Quadrant lawn cutters must be produced per week.
Let the number of Quadrant lawn cutters manufactured in a week be x.
Let the number of Pentagon lawn cutters manufactured in a week be y.
Two of the constraints are:
x > 200
3x + 2y > 2 160
7. Write down the remaining constrain ts in terms of x and y to represent the above men
tioned information.
2. Use graph paper to represent the constraints graphically.
3. Clearly indicate the feasible region by shading it.
4. If the profit on one Quadrant lawn cutter is Rl 200 and the profit on one Pentagon lawn
cutter is R400, write down an equation that will represent the profit on the lawn cutters.
5. Using a search line and your graph, determine the number of Quadrant and Pentagon
lawn cutters that will yield a maximum profit.
6. Determine the maximum profit per week.
SOLUTION
Step I : Remaining constraints:
1
"x+ y <480
i>±
Step 2 : Graphical representation
200
720 960
111
CHAPTER 8. LINEAR PROGRAMMING 8.3
Step 3 : Profit equation
P = 1 200x + 400/
Step 4 : Maximum profit
By moving the search line upwards, we see that the point of maximum profit is
at (600; 900). Therefore
P = 1 200(600) + 400(900)
p = m 080 ooo
Chapter 8
End of Chapter Exercises
1. Polkadots is a small company that makes two types of cards, type X and type Y.
With the available labour and material, the company can make not more than 150
cards of type X and not more than 120 cards of type Y per week. Altogether they
cannot make more than 200 cards per week.
There is an order for at least 40 type X cards and 10 type Y cards per week. Polkadots
makes a profit of R5 for each type X card sold and R10 for each type Y card.
Let the number of type X cards be x and the number of type Y cards be y, manufac
tured per week.
(a) One of the constraint inequalities which represents the restrictions above is x <
150. Write the other constraint inequalities.
(b) Represent the constraints graphically and shade the feasible region.
(c) Write the equation that represents the profit P (the objective function), in terms
of x and y.
(d) On your graph, draw a straight line which will help you to determine how many
of each type must be made weekly to produce the maximum P
(e) Calculate the maximum weekly profit.
2. A brickworks produces "face bricks" and "clinkers". Both types of bricks are pro
duced and sold in batches of a thousand. Face bricks are sold at R150 per thousand,
and clinkers at R100 per thousand, where an income of at least R9 000 per month is
required to cover costs. The brickworks is able to produce at most 40 000 face bricks
and 90 000 clinkers per month, and has transport facilities to deliver at most 100 000
bricks per month. The number of clinkers produced must be at least the same as the
number of face bricks produced.
Let the number of face bricks in thousands be x, and the number of clinkers in
thousands be y,
(a) List all the constraints.
(b) Graph the feasible region.
(c) If the sale of face bricks yields a profit of R25 per thousand and clinkers R45 per
thousand, use your graph to determine the maximum profit.
(d) If the profit margins on face bricks and clinkers are interchanged, use your graph
to determine the maximum profit.
115
8.3 CHAPTER 8. LINEAR PROGRAMMING
3. A small cell phone company makes two types of cell phones: Easyhear and Longtalk.
Production figures are checked weekly. At most, 42 Easyhear and 60 Longtalk phones
can be manufactured each week. At least 30 cell phones must be produced each
week to cover costs. In order not to flood the market, the number of Easyhear phones
cannot be more than twice the number of Longtalk phones. It takes  hour to as
semble an Easyhear phone and  hour to put together a Longtalk phone. The trade
unions only allow for a 50hour week.
Let x be the number of Easyhear phones and y be the number of Longtalk phones
manufactured each week.
(a) Two of the constraints are:
< x < 42 and < y < 60
Write down the other three constraints.
(b) Draw a graph to represent the feasible region
(c) If the profit on an Easyhear phone is R225 and the profit on a Longtalk is R75,
determine the maximum profit per week.
4. Hair for Africa is a firm that specialises in making two kinds of upmarket shampoo,
Glowhair and Longcurls. They must produce at least two cases of Glowhair and one
case of Longcurls per day to stay in the market. Due to a limited supply of chemicals,
they cannot produce more than 8 cases of Glowhair and 6 cases of Longcurls per
day. It takes halfanhour to produce one case of Glowhair and one hour to produce
a case of Longcurls, and due to restrictions by the unions, the plant may operate for
at most 7 hours per day. The workforce at Hair for Africa, which is still in training,
can only produce a maximum of 10 cases of shampoo per day.
Let x be the number of cases of Glowhair and y the number of cases of Longcurls
produced per day.
(a) Write down the inequalities that represent all the constraints.
(b) Sketch the feasible region.
(c) If the profit on a case of Glowhair is R400 and the profit on a case of Longcurls
is R300, determine the maximum profit that Hair for Africa can make per day.
5. A transport contractor has six 5ton trucks and eight 3ton trucks. He must deliver
at least 120 tons of sand per day to a construction site, but he may not deliver more
than 180 tons per day. The 5ton trucks can each make three trips per day at a cost of
R30 per trip, and the 3ton trucks can each make four trips per day at a cost of R120
per trip. How must the contractor utilise his trucks so that he has minimum expense?
More practice (►) video solutions C{J or help at www.everythingmaths.<
(1.)0lgn (2.)01gp (3.)01gq (4.) Olgr (5.)01gs
116
Geometry
9.1 Introduction
In previous years, you learned about the geometry of points, lines and various polygons, made up from
lines. Here we will discuss the geometry of circle in a lot of depth.
© See introductory video: VMhmc at www.everythingmaths.co.za
9.2 Circle Geometry
Terminology
EMCBV
The following is a recap of terms that are regularly used when referring to circles.
arc An arc is a part of the circumference of a circle.
chord A chord is a straight line joining the ends of an arc.
radius A radius, r, is any straight line from the centre of the circle to a point on the circumference.
diameter A diameter, 0, is a special chord that passes through the centre of the circle. A diameter is
the length of a straight line segment from one point on the circumference to another point on
the circumference, that passes through the centre of the circle.
segment A segment is the part of the circle that is cut off by a chord. A chord divides a circle into two
segments.
tangent A tangent is a line that makes contact with a circle at one point on the circumference. (AB is
a tangent to the circle at point P) in Figure 9.1.
Axioms
EMCBW
An axiom is an established or accepted principle. For this section, the following are accepted as
axioms.
117
9.2
CHAPTER 9. GEOMETRY
p tangent
Figure 9.1 : Parts of a circle
Figure 9.2: A rightangled triangle
1. The Theorem of Pythagoras, which states that the square on the hypotenuse of a rightangled
triangle is equal to the sum of the squares on the other two sides. In AABC, this means that
{ABf + (BC) 2 = (AC) 2
2. A tangent is perpendicular to the radius, drawn at the point of contact with the circle.
Theorems of the Geometry of Circles
EMCBX
A theorem is a general proposition that is not selfevident but is proved by reasoning (these proofs need
not be learned for examination purposes).
Theorem 1. The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord.
118
CHAPTER 9. GEOMETRY
9.2
Proof:
Consider a circle, with centre O. Draw a chord AB and draw a perpendicular line from the centre of
the circle to intersect the chord at point P.
The aim is to prove that AP = BP
1 . AOAP and AOBP are rightangled triangles.
2. OA = OB as both of these are radii and OP is common to both triangles.
Apply the Theorem of Pythagoras to each triangle, to get:
OA 2 = OP 2 + AP 2
OB 2
OP 2 + BP 2
However, OA = OB. So,
OP + AP 2
:. AP 2
and AP
OP + BP 2
BP 2
BP
This means that OP bisects AB.
Theorem 2. The line drawn from the centre of a circle, that bisects a chord, is perpendicular to the
chord.
Proof:
Consider a circle, with centre O. Draw a chord AB and draw a line from the centre of the circle to
bisect the chord at point P.
The aim is to prove that OP ± AB
In AOAP and AOBP,
119
9.2
CHAPTER 9. GEOMETRY
1 . AP = PB (given)
2. OA = OB (radii
3. OP is common to both triangles.
AOAP = AOBP (SSS).
OPA
OP A + OPB
:. OPA
:.OP
OPB
180° (APB is a straight line)
OPB = 90°
AB
Theorem 3. The perpendicular bisector of a chord passes through the centre of the circle.
Proof:
Consider a circle. Draw a chord AB. Draw a line PQ perpendicular to AB such that PQ bisects AB
at point P. Draw lines AQ and BQ.
The aim is to prove that Q is the centre of the circle, by showing that AQ = BQ.
In AOAP and AOBP,
1 . AP = PB (given)
2. ZQP.4 = ZQPB (QP _L AB)
3. QP is common to both triangles.
.. AQAP = AQBP (SAS).
From this, QA = QB. Since the centre of a circle is the only point inside a circle that has points on
the circumference at an equal distance from it, Q must be the centre of the circle.
120
CHAPTER 9. GEOMETRY
9.2
Exercise 91
Find the value of x:
2.
5.
f/Vy More practice ( ►) video solutions (*?) or help at www.everythingmaths.co.za
(1.)021i (2.) 021j (3.) 021k (4.) 021m (5.) 021n (6.) 021p
121
9.2
CHAPTER 9. GEOMETRY
Theorem 4. The angle subtended by an arc at the centre of a circle is double the size of the angle
subtended by the same arc at the circumference of the circle.
Proof:
Consider a circle, with centre O and with A and B on the circumference. Draw a chord AB. Draw
radii OA and OB. Select any point P on the circumference of the circle. Draw lines PA and PB.
Draw PO and extend to R.
The aim is to prove that AOB = 2 . APB.
AOR = PAO + APO (exterior angle = sum of interior opp. angles)
But, PAO = APO {AAOP is an isosceles A)
.. AOR = 2APO
Similarly, BOR = 2BPO.
So,
AOB
AOR + BOR
2APO + 2BPO
2(APO + BPO)
2(APB)
122
CHAPTER 9. GEOMETRY
9.2
Exercise 92
Find the angles (a to /) indicated in each diagram:
4. K
fa*) More practice CrJ video solutions Qfj or help at www.everythingmaths.co.za
(1.)021q (2.)021r (3.) 021s (4.) 021t (5.)021u (6.) 021v
Theorem 5. The angles subtended by a chord at the circumference of a circle are equal, if the angles
are on the same side of the chord.
Proof:
12:',
9.2
CHAPTER 9. GEOMETRY
Consider a circle, with centre O. Draw a chord AB. Select any points P and Q on the circumference
of the circle, such that both P and Q are on the same side of the chord. Draw lines PA, PB, QA and
QB.
The aim is to prove that AQB = APB.
AOB 
= 2AQB (Z at centre
and AOB =
= 2APB (Z at centre
.. 2AQB 
= 2APB
:. AQB 
= APB
twice Z at circumference (Theorem 4))
twice Z at circumference (Theorem 4))
Theorem 6. (Converse of Theorem 5) If a line segment subtends equal angles at two other points on
the same side of the line, then these four points lie on a circle.
Proof:
Consider a line segment AB, that subtends equal angles at points P and Q on the same side of AB.
The aim is to prove that points A, B, P and Q lie on the circumference of a circle.
By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let
the circle cut AP (or AP extended) at point R.
AQB
but AQB
.'. ARB
but this cannot be true since ARB
ARB(Zs on same side of chord (Theorem 5))
APB (given)
APB
APB + RBP (exterior Z of A)
.•. the assumption that the circle does not pass through P, must be false, and A, B, P and Q lie on the
circumference of a circle.
Exercise 93
Find the values of the unknown letters.
12 1
CHAPTER 9. GEOMETRY
9.2
Q\*j More practice CwJ video
solutions
C'f) or help at www.everythingmaths.co.za
(1.)021w (2.)021x (3.)021y (4.) 021z (5.) 0220 (6.) 0221
Cyclic Quadrilaterals
Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle. The
vertices of a cyclic quadrilateral are said to be concyclic.
Theorem 7. The opposite angles of a cyclic quadrilateral are supplementary.
Proof:
125
9.2
CHAPTER 9. GEOMETRY
Consider a circle, with centre O. Draw a cyclic quadrilateral ABPQ. Draw AO and PO.
The aim is to prove that ABP + AQP = 180° and QAB + QPB = 180°.
Oi
o 2
But, Oi+Os
.. 2ABP + 2AQP
.. ABP + AQP
Similarly, QAB + QPB
2ABP (Zs at centre (Theorem 4))
2AQP (Zs at centre (Theorem 4))
360°
360°
180°
180°
Theorem 8. (Converse of Theorem 7) If the opposite angles of a quadrilateral are supplementary, then
the quadrilateral is cyclic.
Proof:
Consider a quadrilateral ABPQ, such that ABP + AQP = 180° and QAB + QPB = 180°.
The aim is to prove that points A, B, P and Q lie on the circumference of a circle.
By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let
the circle cut QP (or QP extended) at point R. Draw BR.
QAB + QRB
but QAB + QPB
..QRB
but this cannot be true since QRB
180° (opp. Zs of cyclic quad. (Theorem 7))
180° (given)
QPB
QPB + RBP (exterior Z of A)
120
CHAPTER 9. GEOMETRY
9.2
,\ the assumption that the circle does not pass through P, must be false, and A, B, P and Q lie on the
circumference of a circle and ABPQ is a cyclic quadrilateral.
Exercise 94
Find the values of the unknown letters.
f/Vy More practice (►) video solutions (cj or help at www.everythingmaths.co.za
(1.)0222 (2.) 0223 (3.) 0224 (4.) 0225
Theorem 9. Two tangents drawn to a circle from the same point outside the circle are equal in length.
Proof:
127
9.2
CHAPTER 9. GEOMETRY
Consider a circle, with centre O. Choose a point P outside the circle. Draw two tangents to the circle
from point P, that meet the circle at A and B. Draw lines OA, OB and OP.
The aim is to prove that AP = BP.
In AOAP and AOBP,
1. OA = OB (radii)
2. AOAP = ZOBP = 90° (OA _L AP and OB 1 BP)
3. OP is common to both triangles.
AOAP = AOBP (right angle, hypotenuse, side)
.. AP = BP
Exercise 95
Find the value of the unknown lengths.
128
CHAPTER 9. GEOMETRY
9.2
AE = 5 cm
AC = 8 cm
CE = 9 cm
4.
LN = 7.5 cm
More practice (►) video solutions Cf) or help at www.everythingmaths
(1.)0226 (2.) 0227 (3.) 0228 (4.) 0229
Theorem 10. The angle between a tangent and a chord, drawn at the point of contact of the chord, is
equal to the angle which the chord subtends in the alternate segment.
Proof:
Consider a circle, with centre O. Draw a chord AB and a tangent SR to the circle at point B. Chord
AB subtends angles at points P and Q on the minor and major arcs, respectively.
Draw a diameter BT and join A to T.
The aim is to prove that APB = ABR and AQB = ABS.
First prove that AQB = ABS as this result is needed to prove that APB = ABR.
120
9.2 CHAPTER 9. GEOMETRY
ABS + ABT = 90° (TB ± SR)
BAT = 90° (Zs at centre)
.. ABT + ATB = 90° (sum of angles in ABAT)
:. ABS = ATB
However, AQB = ATB (angles subtended by same chord AB (Theorem 5))
.. AQB = ABS
ABS + ABR = 180° (SBR is a straight line)
APB + AQB = 180° (APBQ is a cyclic quad. (Theorem 7)
From (9.1), AQB = ABS
.. 180°  AQB = 180°  ABS
.. APB = ABR
(9.1)
Exercise 96
Find the values of the unknown letters.
i;so
CHAPTER 9. GEOMETRY
9.2
3. O
131
9.2
CHAPTER 9. GEOMETRY
More practice ( ►) video solutions f?j or help at www.everythingmaths.co.za
(1.)022a (2.) 022b (3.) 022c (4.) 022d (5.) 022e (6.) 022f
(7.) 022g (8.) 022h
Theorem 11. (Converse of 10) If the angle formed between a line, that is drawn through the end point
of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then
the line is a tangent to the circle.
Proof:
Consider a circle, with centre O and chord AB. Let line SR pass through point B. Chord AB subtends
an angle at point Q such that ABS = AQB.
The aim is to prove that SBR is a tangent to the circle.
By contradiction. Assume that SBR is not a tangent to the circle and draw XBY such that XBY is a
tangent to the circle.
ABX
However, ABS
..ABX
But since, ABX
(9.2) can only be true if, XBS
AQB (tanchord theorem)
AQB (given)
ABS
ABS + XBS
(9.2)
If XBS is zero, then both XBY and SBR coincide and SBR is a tangent to the circle.
Exercise 97
1 . Show that Theorem 4 also applies to the following two cases:
1:52
CHAPTER 9. GEOMETRY
9.2
Q\n More practice (►) video solutions ("fj or help at www.everythingmaths.co. z
(1.) 022i
Example 1: Circle Geometry I
QUESTION
BD is a tangent to the circle with centre O.
BO ± AD.
i:«
9.2 CHAPTER 9. GEOMETRY
Prove that:
1. CFOE is a cyclic quadrilateral
2. FB = BC
3. ACOE///ACBF
4. CD 2 = ED x AD
r OE_ _ C_D
BC CO
SOLUTION
Step 1 : To show a quadrilateral is cyclic, we need a pair of opposite angles to be
supplementary, so let's look for that.
FOE = 90° {BO ± OD)
FCE = 90° (Z subtended by diameter AE)
CFOE is a cyclic quadrilateral (opposite Z's supplementary)
Step 2 : Since these two sides are part of a triangle, we are proving that triangle to be
isosceles. The easiest way is to show the angles opposite to those sides to be equal.
Let OEC = x.
FCB = x (Z between tangent BD and chord CE)
BFC = x (exterior Z to cyclic quadrilateral CFOE)
and BF = BC (sides opposite equal Z's in isosceles ABFC)
Step 3 : To show these two triangles similar, we will need 3 equal angles. We already
have 3 of the 6 needed angles from the previous question. We need only find the
missing 3 angles.
CBF = 180°  2x (sum of Z's in ABFC)
OC = OE (radii of circle O)
.. ECO = x (isosceles ACOE)
.. COE = 180°  2x (sum of Z's in ACOE)
• COE = CBF
. ECO = FCB
. OEC = CFB
:. ACOE///ACBF (3 Z's equal)
Step 4 : This relation reminds us of a proportionality relation between similar triangles.
So investigate which triangles contain these sides and prove them similar. In this case 3
equal angles works well. Start with one triangle.
i;;i
CHAPTER 9. GEOMETRY 9.2
In AEDC
CED = 180°  x (Z's on a straight line AD)
ECD = 90° — x (complementary Z's)
Step 5 : Now look at the angles in the other triangle.
In AADC
ACT) = 180°  x (sum of Z's ACE and ECD)
CAD = 90°  x (sum of Z's in ACAE)
Step 6 : The third equal angle is an angle both triangles have in common.
Lastly, ADC = EDC since they are the same Z.
Step 7 : Now we know that the triangles are similar and can use the proportionality
relation accordingly.
:. AADC 1 1 1 ACDE (3 Z's equal)
ED _ CD
' ' ~CD ~ AD
:. CD 2 = ED x AD
Step 8 : This looks like another proportionality relation with a little twist, since not all
sides are contained in 2 triangles. There is a quick observation we can make about the
odd side out, OE.
OE = OC (AOEC is isosceles)
Step 9 : With this observation we can limit ourselves to proving triangles BOC and
ODC similar. Start in one of the triangles.
In ABCO
OCB = 90° (radius OC on tangent BD)
CBO = 180°  2x (sum of Z's in ABFC)
BOC = 2x 90° (sum of Z's in ABCO)
Step 1 : Then we move on to the other one.
In AOCD
OCD = 90° (radius OC on tangent BD)
COD = 180°  2x (sum of Z's in AOCE)
CDO = 2x 90° (sum of Z's in AOCD)
Step 1 1 : Then, once we've shown similarity, we use the proportionality relation, as
well as our first observation, appropriately.
ABOC 1 1 1 AODC (3 Z's equal)
CO _ CD
BC ~ CO
OF 1 CD
^— = ^— (OE = CO isosceles AOEC)
BC CO
135
9.2
CHAPTER 9. GEOMETRY
Example 2: Circle Geometry II
QUESTION
FD is drawn parallel to the tangent CB
Prove that:
1. FADE is cyclic
2. AAFE///ACBD
3 FCxAG _ DCxFE
SOLUTION
Step 7 : In this case, the best way to show FADE is a cyclic quadrilateral is to look for
equal angles, subtended by the same chord.
Let /LBCD = x
:. AC AH = x (Z between tangent BC and chord CE)
.. Z_FDC = x (alternate Z, FD  CB)
.. FADE is a cyclic quadrilateral (chord FE subtends equal Z's)
Step 2 : To show these 2 triangles similar we will need 3 equal angles. We can use the
result from the previous question.
i;s(i
CHAPTER 9. GEOMETRY 9.2
Let ZFEA = y
:. ZFDA = y (Z's subtended by same chord AF in cyclic quadrilateral FADE)
.. ZCBD = y (corresponding Z's, FD \\ CB)
.. ZFEA = ZCBD
Step 3 : We have already proved 1 pair of angles equal in the previous question.
ZBCD = ZFAE (above)
Step 4 : Proving the last set of angles equal is simply a matter of adding up the angles
in the triangles. Then we have proved similarity.
ZAFE = 180° xy (Z's in AAFE)
ZCDB = 180° xy (Z's in ACBD)
:. AAFE///ACBD (3 Z's equal)
Step 5 : This equation looks like it has to do with proportionality relation of similar
triangles. We already showed triangles AFE and CBD similar in the previous question.
So lets start there.
DC FA
KD FE
DC x FE
BD
FA
Step 6 : Now we need to look for a hint about side FA. Looking at triangle CAH we
see that there is a line FG intersecting it parallel to base CH. This gives us another
proportionality relation.
AC FA
p— = — =; (FG  CH splits up lines AH and AC proportionally)
GH r C
^ A FCxAG
■• FA =^H—
Step 7 : We have 2 expressions for the side FA.
FC.AG _ DC x FE
' ' GH ~ BD
137
9.3
CHAPTER 9. GEOMETRY
9.3 Coordinate Geometry
Equation of a Circle
EMCBZ
We know that every point on the circumference of a circle is the same distance away from the centre of
the circle. Consider a point (xi;yi) on the circumference of a circle of radius r with centre at (xo;yo).
\P(xi;yi)
Figure 9.3: Circle with centre (x ;yo) and a point P at (x\;yi)
In Figure 9.3, AOPQ is a rightangled triangle. Therefore, from the Theorem of Pythagoras, we know
that:
But,
OP 2 = PQ 2 + OQ 2
PQ = Vi yo
OQ = xi — xo
OP = r
r 2 = (yi  yo) 2 + (si  x ) 2
But, this same relation holds for any point P on the circumference. Therefore, we can write:
(xxo) 2 + (y yo) 2 = r 1
for a circle with centre at (xo\ yo) and radius r.
For example, the equation of a circle with centre (0;0) and radius 4 is:
(y  yo) 2 + (x x ) 2 = r 2
(y0) 2 + (x0) 2 = 4 2
16
2 . 2
x +y
(9.3)
© See video: VMhrm at www.everythingmaths.co.za
l:S.s
CHAPTER 9. GEOMETRY
Example 3: Equation of a Circle I
QUESTION
Find the equation of a circle (centre O) with a diameter between two points, P at (—5; 5) and
Qat(5;5).
SOLUTION
Step 1 : Draw a picture
Draw a picture of the situation to help you figure out what needs to be done.
Step 2
Find the centre of the circle
We know that the centre of a circle lies on the midpoint of a diameter. Therefore
the coordinates of the centre of the circle is found by finding the midpoint of the
line between P and Q. Let the coordinates of the centre of the circle be (x ;yo),
let the coordinates of P be (xi;yi) and let the coordinates of Q be (x 2 ;y2)
Then, the coordinates of the midpoint are:
x
Vo
xi + x 2
2
5 + 5
2
m + y2
2
5 + (5)
2
=
The centre point of line PQ and therefore the centre of the circle is at (0; 0).
Step 3 : Find the radius of the circle
If P and Q are two points on a diameter, then the radius is half the distance
between them.
139
CHAPTER 9. GEOMETRY
The distance between the two points is:
 2 PQ =
7jV( X 2 Xl) 2 + (2/2  J/l) 2
=
2 V(5(5))H(55^
=
2V(10) 2 + (10) 2
1
=
Vioo + ioo
J200
V 4
V50
Step 4 : Write the equation of the circle
x + y =50
Example 4: Equation of a Circle II
QUESTION
Find the centre and radius of the circle
x 2  Ux + y 2 +Ay = 28.
SOLUTION
Step 1 : Change to standard form
We need to rewrite the equation in the form (x — xq) + {y — j/o) = r
To do this we need to complete the square
i.e. add and subtract ( cooefficient of x) 2 and ( cooefficient of y) 2
Step 2 : Adding cooefficients
x 2  lAx + y 2 + 4y = 28
.. x 2  Ux + (7) 2  (7) 2 + y 2 + Ay + (2) 2  (2) 2 = 28
Step 3 : Complete the squares
:. (i7) 2 (7) 2 + (2/ + 2) 2 (2) 2 = 28
Step 4 : Take the constants to the other side
.. (i7) 2 49 + (j/ + 2) 2 4 = 28
.. {x  7) 2 + (y + 2) 2 = 28 + 49 + 4
.. (x7) 2 + (y + 2) 2 =25
Step 5 : Read the values from the equation
110
CHAPTER 9. GEOMETRY 9.3
centre is (7; —2) and the radius is 5 units
Equation of a Tangent to a Circle at a Point ^emcca
on the Circle
We are given that a tangent to a circle is drawn through a point P with coordinates (xi;yi). In this
section, we find out how to determine the equation of that tangent.
(a>i ;ift)
Figure 9.4: Circle h with centre (x ;yo) has a tangent, g passing through point P at (xi;yi). Line /
passes through the centre and point P.
We start by making a list of what we know:
1. We know that the equation of the circle with centre (xo;yo) and radius r is (2 — x ) 2 + (y— yo) 2 =
r 2 .
2. We know that a tangent is perpendicular to the radius, drawn at the point of contact with the
circle.
As we have seen in earlier grades, there are two steps to determining the equation of a straight line:
Step 1 : Calculate the gradient of the line, m.
Step 2: Calculate the ^intercept of the line, c.
The same method is used to determine the equation of the tangent. First we need to find the gradient
of the tangent. We do this by finding the gradient of the line that passes through the centre of the circle
and point P (line / in Figure 9.4), because this line is a radius and the tangent is perpendicular to it.
2/i  Vo ,„ .,
mr = (9.4)
Xi  xo
The tangent (line g) is perpendicular to this line. Therefore,
rrif X m 9 = —1
So,
1
m„ =
mj
141
9.3
CHAPTER 9. GEOMETRY
Now, we know that the tangent passes through (xi; yi) so the equation is given by:
;  xi)
(x  xi)
y  2/1 = m g (x — xi)
1
y 2/i =
771/
y i/l yiyo * 1 '
xixo
Xi  x ,
yyi
yi  2/0
1 / N
(x  Xl)
For example, find the equation of the tangent to the circle at point (1: 1). The centre of the circle is at
(0; 0). The equation of the circle is x 2 + y 2 = 2.
Use
2/ 2/i
xi  x a
2/i  2/o
(x  Xi)
with(xo;j/o) = (0:0)and(xi;j/i) = (1; 1).
2/ 2/1
—
: (x
2/1 2/o
2/1
=
(x — 1)
10 V '
2/1
=
j(*"l)
y
=
(xl) + l
y
=
x + 1 + 1
V
=
x + 2
Exercise 98
1. Find
(a)
(b)
(0
(d)
(e)
2. (a)
(b)
(c)
3. (a)
(b)
(0
4. Find
(a)
(b)
(0
(d)
(e)
the equation of the circle:
with centre (0; 5) and radius 5
with centre (2; 0) and radius 4
with centre (5; 7) and radius 18
with centre (—2: 0) and radius 6
with centre (—5; —3) and radius \/3
Find the equation of the circle with centre (2; 1) which passes through (4; 1).
Where does it cut the line y = x + 1?
Draw a sketch to illustrate your answers.
Find the equation of the circle with centre (—3: —2) which passes through (1; —4).
Find the equation of the circle with centre (3; 1) which passes through (2; 5).
Find the point where these two circles cut each other.
the centre and radius of the following circles:
(x  9) 2 + (y 6) 2 = 36
(x  2) 2 + (y  9) 2 = 1
(x + 5) 2 + (y + 7) 2 = 12
(x + 4) 2 + (y + 4) 2 = 23
3(x2) 2 + 3(i/ + 3) 2 = 12
1 12
CHAPTER 9. GEOMETRY
9.4
(f) x 2  3x + 9 = y 2 + 5y + 25 = 17
5. Find the re and y intercepts of the following graphs and draw a sketch to illustrate your answer:
(a) (x + 7) 2 + (y2) 2 = 8
(b) x 2 + (y  6) 2 = 100
(c) {x + if + y 2 = 16
(d) {x  5) 2 + (y + l) 2 = 25
6. Find the centre and radius of the following circles:
(a) x 2 + 6x + y 2  12y = 20
(b) x 2 + 4x + y 2  %y =
(c) x 2 + y 2 + % = 7
(d) x 2  Qx + y 2 = 16
(e) x 2  5x + y 2 + 3y = 
(f) x 2  %nx + y 2 + lOny = 9n 2
7. Find the equation of the tangent to each circle:
(a) x 2 +y 2 = 17 at the point (1; 4)
(b) x 2 +y 2 = 25 at the point (3; 4)
(c) (x + l) 2 + (y  2) 2 = 25 at the point (3; 5)
(d) (x  2) 2 + (y l) 2 = 13 at the point (5; 3)
f/Vy More practice f ►) video solutions f Oj or help at www.everythingmaths.co.za
(1.)01gt (2.)01gu (3.)01gv (4.)01gw (5.)01gx (6.) 01gy
(7.)01gz
9.4 Transformations
Rotation of a Point About an Angle 9
EMCCC
First we will find a formula for the coordinates of P after a rotation of (
We need to know something about polar coordinates and compound angles before we start.
113
CHAPTER 9. GEOMETRY
Polar coordinates
Notice that sin a =  : .y = r sin a
and cos a = — .'. x = r cos a
so P can be expressed in two ways:
P(x:y) rectangular coordinates
or P(r; a) polar coordinates.
Compound angles
(See derivation of formulae in Chapter 12)
sin (a + f3)
cos (a + f3)
sin a cos f) + sin /3 cos a
cos a cos P — sin a sin /?
Now consider P' after a rotation of (
P(x; y) = P(r cos ct; r sin a)
P (r cos (a + #);r sin (a + 8))
Expand the coordinates of P'
coordinate
y — coordinate
rcos(a + 8)
r [cos a cos 8 — sin a sin 6]
r cos a cos 6 — r sin a sin #
x cos 8 — y sin 6
r sin (a + 6)
r [sin a cos # + sin 8 cos a]
r sin a cos 8 + r cos a sin
i/ cos 8 + x sin #
which gives the formula P' = [(x cos 8 — y sin 8; y cos 6 + x sin #)]
p'
e
a \
P =
(rcosa;
So to find the coordinates of P(l; %/3) after a rotation of 45°, we arrive at:
P = [(xcosd — ysm8);(ycos8 + xs'mS)]
= [(lcos45°  V3sin45°);(V3cos45° + 1 sin 45°)
,\/2 v/2/'\V2 V2
1 V3 y^+1
^2 ' ^2
Exercise 99
114
CHAPTER 9. GEOMETRY
9.4
Any line OP is drawn (not necessarily in the first
quadrant), making an angle of 9 degrees with the
xaxis. Using the coordinates of P and the angle
a, calculate the coordinates of P', if the line OP
is rotated about the origin through a degrees.
P
a
1.
(2; 6)
60°
2.
(4; 2)
30°
3.
(5;i)
45°
4.
(3; 2)
120°
5.
(4;i)
225°
6.
(2; 5)
150°
o
More practice ( ►) video solutions (cj or help at www.everythingmaths.co.za
(1.)022j
Characteristics of Transformations
EMCCD
Rigid transformations like translations, reflections, rotations and glide reflections preserve shape and
size, and that enlargement preserves shape but not size.
Activity:
Geometric Transformations
Draw a large 15x15 grid and plot
AABC with A(2;6), B(5;6) and
C(5; 1). Fill in the lines y = x and
y = x.
Complete the table below , by draw
ing the images of AABC under the
given transformations. The first one
has been done for you.
15
10
5
A(2;6) B(5;6)'
7(5; 1)
15
10
' y = x
—5
10
5 10
'•., *„,
'.'I
/ 1
*  I
A' B'
y = I ■
15
15
115
9.4
CHAPTER 9. GEOMETRY
Transformation
Description
(translation, reflection,
rotation, enlargement)
Coordinates
Lengths
Angles
(x;y) > (x;y)
reflection about the xaxis
A' (2; 6)
B'(5;6)
C"(5;2)
A'B' = 3
B'C*' = 4
A'C = 5
B' = 90°
tan A = 4/3
.. i = 53°,C , = 37°
(x;y) »■ (x + 1; 1/  2)
(a?;y) > (a;;i/)
(x;y) > (i/;x)
(ar;y) ► (a; y)
(x;y) + (2x;2y)
(x;y) * (y;x)
(x;y) t(y;x + 1)
A transformation that leaves lengths and angles unchanged is called a rigid transformation.
Which of the above transformations are rigid?
Chapter 9
End of Chapter Exercises
1 . AABC undergoes several transformations forming AA'B'C Describe the relation
ship between the angles and sides of AABC and AA'B'C' (e.g., they are twice as
large, the same, etc.)
Transformation
Sides
Angles
Area
Reflect
Reduce by a scale factor of 3
Rotate by 90°
Translate 4 units right
Enlarge by a scale factor of 2
2. ADEFhasE = 30°, DE = 4 cm, EF = 5 cm. ADEF is enlarged by a scale factor
of 6 to form AD'E'F'.
(a) Solve ADEF
(b) Hence, solve AD'E'F'
3. AXYZ has an area of 6cm 2 . Find the area of AX'Y'Z' if the points have been
transformed as follows:
(a) (x,y)^(x + 2;y + 3)
(b) (x,y) > (y;x)
ll(i
CHAPTER 9. GEOMETRY 9.4
(c) (a;
,y)
► (4x:
J/)
(d) (x
,y)
+ (3a;:
2/ + 2)
(e) (x
,y)
— > (—a
•;?/)
0) (x
,v)
► (a;;
2/ + 3)
(g) (a:
,y)
> (4s;
%)
(h) (x
,y)
^(3x;4y)
ft" 1 ) More practice (►) video solutions f9) or help at www.everythingmaths
(1.) OlhO (2.) Olhl (3.)01h2
147
Trigonometry
7 0. 1 Introduction
In this chapter we will build on the trigonometry from previous years by looking at the result applying
trigonometric functions to sums and differences of angles. We will also apply trigonometry to solving
geometric problems in two and three dimensions.
© See introductory video: VMhtu at www.everythingmaths.co.za
10.2 Compound Angle Identities
Derivation of sin (a + (5)
EMCCG
We have, for any angles a and 0, that
sin(a + 0) = sin a cos + sin /3 cos a
How do we derive this identity? It is tricky, so follow closely.
Suppose we have the unit circle shown below. The two points L(a; b) and K(x; y) are on the circle.
148
CHAPTER 10. TRIGONOMETRY
10.2
K(x;y)
L(a:b)
We can get the coordinates of L and K in terms of the angles a and 0. For the triangle LOM, we have
that
sin /3 = —
a a
cos p = —
b = sin /3
a = cos /3
Thus the coordinates of L are (cos /3; sin /3). In the same way as above, we can see that the coordinates
of K are (cos a; sin a). The identity for cos(a — /3) is now determined by calculating KI? in two ways.
Using the distance formula (i.e. d = \J\x~i — ii) 2 + (2/2 — 3/1) 2 or d 2 = (x 2 — xi) 2 + (y 2 — yi) 2 ), we
can find KL 2 :
KL = (cos a — cos/3) + (sin a — sin/3)
= cos a — 2 cos a cos /3 + cos /3 + sin a — 2 sin a sin /3 + sin /3
= (cos a + sin a) + (cos /3 + sin /3) — 2 cos a cos /3 — 2 sin a sin /3
= 1 + 1 — 2(cos a cos /3 + sin a sin /3)
= 2 — 2(cosacos/3 + sinasin/3)
The second way we can determine KL 2 is by using the cosine rule for AKOL:
KL 2 = KO 2 + LO 2 2.KO.LO. cos(a  0)
= 1 2 + l 2 2(l)(l)cos(a/3)
= 2  2 . cos(a  13)
Equating our two values for KL 2 , we have
2 — 2 . cos(a — /3) = 2 — 2(cosacos/3 + sin a . sin/3)
^=> cos(a — /3) = cos a . cos /3 + sin a . sin /3
Now let a — > 90° — a. Then
cos(90°  a  /3) = cos(90°  a) cos /3 + sin(90° a)sin/3
= sin a . cos /3 + cos a . sin /3
But cos(90°  (a + /3)) = sin(a + /3). Thus
sin(a + /3) = sin a . cos /3 + cos a . sin /3
J. J.<)
70.2
CHAPTER 10. TRIGONOMETRY
Derivation of sin (a  (3)
EMCCH
We can use
to show that
We know that
and
Therefore,
sin(a + 0) = sin a cos + cos a sin
sin(a — 0) = sin a cos — cos a sin
sin(fl) = sin(0)
cos(— 0) = COS0
sin(a  /3) = sin(a + (£))
= sinacos(— 0) + cosasin(— /3)
= sin a cos /3 — cos a sin /3
Derivation of cos (a + /5)
EMCCI
We can use
to show that
We know that
Therefore,
sin(a — 0) = sin a cos — sin cos a
cos(a + 0) = cos a cos — sin a sin
sin(fl) = cos(906»).
cos(a + /3) = sin(90 (a + /3))
= sin((90  a)  0))
= sin(90 — a) cos/3 — sin/3cos(90 — a)
= cos a cos — sin /3 sin a
Derivation of cos (a  /3)
EMCCJ
We found this identity in our derivation of the sin(o + 0) identity. We can also use the fact that
sin(ct + 0) = sin a cos + cos a sin
to derive that
As
cos(ct — 0) = cos a cos + sin a sin
cos(0) = sin(906»),
150
CHAPTER 10. TRIGONOMETRY
10.2
we have that
cos(ct P) = sin(90  (a  /?))
= sin((90  a) + /?))
= sin(90 — q) cos + cos(90 — a) sin /3
= cos a cos f) + sin a sin /3
Derivation of sin 2a
EMCCK
We know that
When a = /9, we have that
sin(a + /3) = sin « cos /3 + cos a sin /3
sin(2a) = sin(a + a) = sin a cos a + cos a sin a
= 2 sin a cos a
Derivation of cos 2a
EMCCL
We know that
When a = /}, we have that
cos(a + ft) = cos a cos /3 — sin a sin /3
cos(2a) = cos(q + a) = cos a cos a — sin a sin a
cos a — sin a
However, we can also write
and
by using
Activity:
cos 2o = 2 cos a — 1
cos 2a = 1 — 2 sin a
sin a + cos a = 1 .
77ie cos 2a Identity
Use
to show that:
151
70.2
CHAPTER 10. TRIGONOMETRY
Problemsolving Strategy for Identities
EMCCM
Tip
When
proving trigono
metric
identities, never
assume that the [eft hand
side is
equal to the right
hand side. You need to
show that both sides are
equal.
The most important thing to remember when asked to prove identities is:
A suggestion for proving identities: It is usually much easier simplifying the more complex side of an
identity to get the simpler side than the other way round.
Example 1: Trigonometric Identities 1
QUESTION
Prove that sin 75° = (v ^ 3+1) without using a calculator.
SOLUTION
Step 1 : Identify a strategy
We only know the exact values of the trig functions for a few special angles (30°;
45°; 60°; etc.). We
can see that 75° = 30° + 45°. Thus we can use our double
angle identity for sin(a + 0) to express sin 75° in terms of known trig function
values.
Step 2 : Execute strategy
sin 75° =
sin(45°+30°)
=
sin(45°) cos(30°) + cos(45°) sin(30°)
=
1 ^3 1 1
v/2' 2 + ^2'2
73 + 1
2^2
n/3 + 1 V2
2s/2 '" s/2
v/2(\/3 + l)
4
lo2
CHAPTER 10. TRIGONOMETRY 10.2
Example 2: Trigonometric Identities 2
QUESTION
Deduce a formula tor tan(a + 0) in terms of tana and tan/3.
Hint: Use the formulae tor sin(a + 0) and cos(a + 0)
SOLUTION
Step 1 : Identify a strategy
We can express tan(a + 0) in terms of cosines and sines, and then use the
doubleangle formulae for these. We then manipulate the resulting expression in
order to get it in terms of tana and tan/3.
Step 2 : Execute strategy
tan(a + /3)
sin(a + /3)
cos(a + /3)
sin a . cos + cos a . sin /3
cos a . cos — sin a . sin
sin a . cos j3 ■ cos a . sin /3
cos a . cos cos a . cos /3
tan a + tan
1 — tana . tan/3
Example 3: Trigonometric Identities 3
QUESTION
Prove that
sin 6 + sin 20
—  = tan
1 + cos 9 + cos 20
In fact, this identity is not valid for all values of 9. Which values are those?
SOLUTION
Step 1 : Identify a strategy
The righthand side (RHS) of the identity cannot be simplified. Thus we should
try simplify the lefthand side (LHS). We can also notice that the trig function on
the RHS does not have a 29 dependence. Thus we will need to use the double
angle formulae to simplify the sin 20 and cos 29 on the LHS. We know that tan#
l.VS
70.2 CHAPTER 10. TRIGONOMETRY
is undefined for some angles 9. Thus the identity is also undefined for these 9,
and hence is not valid for these angles. Also, for some 9, we might have division
by zero in the LHS, which is not allowed. Thus the identity won't hold for these
angles also.
Step 2 : Execute the strategy
sin 9 + 2 sin 9 cos I
LHS
1 + cos 9 + (2 cos 2 9  1)
sin6»(l + 2cos#)
cos6»(l + 2cos6>)
sin 9
cos 9
tan 9
RHS
We know that tan 6 is undefined when 9 = 90° + 180°n, where n is an integer.
The LHS is undefined when 1 + cos 9 + cos 29 = 0. Thus we need to solve this
equation.
1 + cos 9 + cos 29 =
=>• cos6»(l + 2cos6») =
The above has solutions when cos9 = 0, which occurs when 9 = 90° + 180°n,
where n is an integer. These are the same values when tan# is undefined. It
also has solutions when 1 + 2 cos 9 = 0. This is true when cos 9 = — \, and thus
9 = ...  240°, 120°, 120°, 240°, .... To summarise, the identity is not valid
when 9 = ...  270° ; 240° ;  120° ; 90° ; 90° ; 120° ; 240° ; 270° ; . . .
Example 4: Trigonometric Equations
QUESTION
Solve the following equation for y without using a calculator:
1 — sin y — cos 2y
sin 2y — cos y
SOLUTION
Step 1 : Identify a strategy
Before we are able to solve the equation, we first need to simplify the lefthand
side. We do this by using the doubleangle formulae.
Step 2 : Execute the strategy
lol
CHAPTER 10. TRIGONOMETRY
10.3
1 — sin y — (1 — 2 sin y)
2 sin y cos i/ — cos y
2 sin y — sin y
cos 1/(2 sin y — 1)
sinj/(2sini/ — 1)
cosy(2sinj/ — 1)
tan y
/ = 135° + 180° n; n e Z
1
1
1
1
10.3 Applications of Trigonometric
Functions
Problems in Two Dimensions
EMCCO
Example 5: Problem in Two Dimensions
QUESTION
A
o 1
I °
X / N^~~~\^^
c *^ — ■
B
Tor the figure below, we are given that
CD = BD = x.
Show that BC 2 = 2x 2 ( 1 + sin 6) .
ir>r>
70.3
CHAPTER 10. TRIGONOMETRY
SOLUTION
Step 7 : Identify a strategy
We want CB, and we have CD and BD. If we could get the angle BDC,
then we could use the cosine rule to determine BC. This is possible, as AABD
is a rightangled triangle. We know this from circle geometry, that any triangle
circumscribed by a circle with one side going through the origin, is rightangled.
As we have two angles of AABD, we know ADB and hence BDC. Using the
cosine rule, we can get BC 2 .
Step 2 : Execute the strategy
Thus
ADB = 180°
BDC
90
: 90°  e
180°  ADB
= 180°  (90°  0)
= 90° +
Now the cosine rule gives
BC 2 = CD 2 + BD 2 2.CD.BD. cos(BDC)
= x 2 + x 2  2 . x 2 . cos(90° + 0)
= 2x 2 + 2x 2 [sin(90°) cos(0) + sin(0) cos(90°)]
= 2x 2 + 2x 2 [ 1 . cos(0) + sin(6>) . 0]
= 2x 2 (lsin6»)
Exercise 101
1 . For the diagram on the right,
l"i(i
CHAPTER 10. TRIGONOMETRY
10.3
(a) Find AOC in terms of 9
(b) Find an expression for:
cosy
sin 9
sin 26»
(c) Using the above, show that sin 2$
2 sin 9 cos 9.
(d) Now do the same for cos 29 and tan 6.
2. DC is a diameter of circle O with radius r. CA = r, AB = DE and DOE ■
Show that cos 9 = ,
E
3. The figure below shows a cyclic quadrilateral with ^ = ^.
(a) Show that the area of the cyclic quadrilateral \s DC .DA. sinD.
(b) Find expressions for cosD and cosB in terms of the quadrilateral sides.
(c) Show that 2CA 2 = CD 2 + DA 2 + AB 2 + BC 2 .
(d) Suppose that BC = 10, CD = 15, AD = 4 and AB = 6. Find CA 2 .
(e) Find the angle D using your expression for cos D. Hence find the area of ABCD.
Q^j More practice f ►) video solutions (7) or help at www.everythingmaths.co.za
(l.)01ai (2.)01aj (3.) Olak
157
70.3
CHAPTER 10. TRIGONOMETRY
Problems in 3 Dimensions
EMCCP
Example 6: Height of tower
QUESTION
D is the top of a tower of height h. Its base is at A. The triangle ABC lies on the ground (a
horizontal plane). If we have that BC = b, DBA = a, DBC = and DCB = 9, show that
h
b sin q sin 9
sin(/3 + 9)
SOLUTION
1").S
CHAPTER 10. TRIGONOMETRY 10.3
Step 1 : Identify a strategy
We have that the triangle ABD is rightangled. Thus we can relate the height
h with the angle a and either the length BA or BD (using sines or cosines).
But we have two angles and a length for ABCD, and thus can work out all the
remaining lengths and angles of this triangle. We can thus work out BD.
Step 2 : Execute the strategy
We have that
h
=— = sin a
BD
=*> h = BDsvnct
Now we need BD in terms of the given angles and length b. Considering the
triangle BCD, we see that we can use the sine rule.
sin sm(BDC)
~BD 6
BD
But BDC = 180° 139, and
bsmd
sm{BDC)
sin(180° /30) =  sin(/3 
= sin(/3 + &)
So
BD
b sin 6
sm(BDC)
bsmd
sin(/3 + 9)
BD sin a
b sin a sin 9
sin(/3 + 9)
Exercise 102
1 . The line BC represents a tall tower, with B at its foot. Its angle of elevation from D is 9. We are
also given that BA = AD = x.
159
70.4
CHAPTER 10. TRIGONOMETRY
C
D
2. Find the height of the tower BC in terms of x, tan 61 and cos 2a.
3. Find BC if we are given that x = 140 m, a = 21° and 9 = 9°.
I\*j More practice f ►) video solutions CfJ or help at www.everythingmaths.co.za
(1.)01am (2.)01an (3.) 01ap
10.4 Other Geometries
Taxi cab Geometry
EMCCR
Taxicab geometry, considered by Hermann Minkowski in the 19th century, is a form of geometry in
which the usual metric of Euclidean geometry is replaced by a new metric in which the distance
between two points is the sum of the (absolute) differences of their coordinates.
Manhattan Distance
EMCCS
The metric in taxicab geometry, is known as the Manhattan distance, between two points in an
Euclidean space with fixed Cartesian coordinate system as the sum of the lengths of the projections of
the line segment between the points onto the coordinate axes.
Kid
CHAPTER 10. TRIGONOMETRY
10.4
For example, the Manhattan distance between the point Pi with coordinates (xi;yi) and the point P 2
at {xr,yi) is
xi x 2 \ + \yi yi\
(10.1)
' • • • • ■ • i m • • • *~*~i
! I
• >
, ,
Figure 10.1: Manhattan distance (dotted and solid) compared to Euclidean distance (dashed). In each
case the Manhattan distance is 12 units, while the Euclidean distance is a/36
The Manhattan distance changes if the coordinate system is rotated, but does not depend on the
translation of the coordinate system or its reflection with respect to a coordinate axis.
Manhattan distance is also known as city block distance or taxicab distance. It is given these names
because it is the shortest distance a car would drive in a city laid out in square blocks.
Taxicab geometry satisfies all of Euclid's axioms except for the sideangleside axiom, as one can
generate two triangles with two sides and the angle between them the same and have them not be
congruent. In particular, the parallel postulate holds.
A circle in taxicab geometry consists of those points that are a fixed Manhattan distance from the
centre. These circles are squares whose sides make a 45° angle with the coordinate axes.
161
70.4
CHAPTER 10. TRIGONOMETRY
Summary of the Trigonometric
Rules and Identities
Pythagorean Identity
cos 2 9 + sin 2 9 = 1
Cofunction Identities
sin(90°  8) = cos 9
cos(90° 8) = sin8
Ratio Identities
tan 8 ■
sin 9
cos
Odd/Even Identities
sin(— 8) = — sin 8
cos(—8) = cos 8
tan(— 8) = —tan 8
Periodicity Identities Double Angle Identities
sin(6»±360°) = sinfl
cos(0±36O°) = cos 6»
tan(6>±180°) = tan6»
sm(28) = 2 sin0 cos 8
cos (28) = cos 2 8 — sin 2
cos (261) = 1 2sin 2
tan (26))= 1 a '" l °
Addition/Subtraction Identities
Area Rule
Cosine rule
sin(6» + ^
sin (8 — <j>
cos (9 + 4>
cos (8 — <f>
tan (8 + <p
tan (8 — (j)
■ sin U cos <p + cos o sin q>
■ sin 8 cos </> — cos 8 sin
= cos 8 cos </> — sin 8 sin
= cos 8 cos + sin 8 sin </>
tan <p+tan
1— tan 6 tan ci
tan <A — tan
Area = \bc sin A
Area =  ah sin C
Area = \ac sin B
a 2 = 6 2 + c 2  2bccosA
6 2 = a 2 + c 2  2ac cos B
c 2 = a 2 + b 2 2abcosC
1+tan 8 tan <f>
Sine Rule
Chapter 10
End of Chapter Exercises
Do the following without using a calculator.
1. Suppose cosd = 0,7. Find cos 29 and cos id.
2. If sin 8 = , again find cos 29 and cos 46 1 .
3. Work out the following:
(a) cos 15°
(b) cos 75°
(c) tan 105°
(d) cos 15°
(e) cos 3° cos 42° — sin 3° sin 42°
(f) l2sin 2 (22,5°)
4. Solve the following equations:
(a) cos 39 . cos 9 — sin 39 . sin 9 = — \
11,2
CHAPTER 10. TRIGONOMETRY 10.4
(b) 3 sin 8 = 2cos 2 6>
5. Prove the following identities
(a) Sin 3 6 = 3 si 9 sin 38
(b) cos 2 q(1 — tan 2 a) = cos la
(c) 4 sin 8 . cos 8 . cos 26* = sin 48
(d) 4 cos 3 x — 3 cos x = cos 3x
/ \ i sin lit
(e) tan !/ = — „ , .
6. (Challenge question!) If a + b + c = 180°, prove that
sin a+sin 6+sin c = 3cos(a/2) cos(6/2) cos(c/2)+cos(3a/2) cos(36/2) cos(3c/2)
A"y More practice f ►) video solutions \jf) or help at www.everythingmaths.co.za
(1.)01aq (2.)01ar (3.) 01as (4.) 01 at (5.) Olau (6.) 01av
l(i:i
Statistics
1 1 .1 Introduction
EMCCU
In this chapter, you will use the mean, median, mode and standard deviation of a set of data to identify
whether the data is normally distributed or whether it is skewed. You will learn more about populations
and selecting different kinds of samples in order to avoid bias. You will work with lines of best fit, and
learn how to find a regression equation and a correlation coefficient. You will analyse these measures
in order to draw conclusions and make predictions.
© See introductory video: VMhwi at www.everythingmaths.co.za
7 7.2 Normal Distribution
Activity:
You are given a table of data below.
75
67
70
71
71
73
74
75
80
75
77
78
78
78
78
79
91
81
82
82
83
86
86
87
1 . Calculate the mean, median, mode and standard deviation of the data.
2. What percentage of the data is within one standard deviation of the mean?
3. Draw a histogram of the data using intervals 60 < x < 64; 64 < x < 68; etc.
4. Join the midpoints of the bars to form a frequency polygon.
If large numbers of data are collected from a population, the graph will often have a bell shape. If the
data was, say, examination results, a few learners usually get very high marks, a few very low marks
and most get a mark in the middle range. We say a distribution is normal if
the mean, median and mode are equal.
it is symmetric around the mean.
±68% of the sample lies within one standard deviation of the mean, 95% within two standard
deviations and 99% within three standard deviations of the mean.
164
CHAPTER 11. STATISTICS
11.2
x + a x + 2cr x + 3a
What happens if the test was very easy or very difficult? Then the distribution may not be symmetrical.
If extremely high or extremely low scores are added to a distribution, then the mean tends to shift
towards these scores and the curve becomes skewed.
If the test was very difficult, the mean score is shifted to the
left. In this case, we say the distribution is positively skewed,
or skewed right.
If it was very easy, then many learners would get high scores,
and the mean of the distribution would be shifted to the right.
We say the distribution is negatively skewed, or skewed left.
Skewed right
Skewed left
Exercise 111
1. Given the pairs of normal curves below, sketch the graphs on the same set of axes and show
any relation between them. An important point to remember is that the area beneath the curve
corresponds to 100%.
(a) Mean = 8, standard deviation = 4 and Mean
(b) Mean = 8, standard deviation = 4 and Mean
(c) Mean = 8, standard deviation = 4 and Mean
4, standard deviation =
16, standard deviation
8, standard deviation =
8
2. After a class test, the following scores were recorded:
Test Score
Frequency
3
1
4
7
5
14
6
21
7
14
8
6
9
1
Total
64
Mean
6
Standard Deviation
1,2
(a) Draw the histogram of the results.
(b) Join the midpoints of each bar and draw a frequency polygon.
lfi.")
11.3
CHAPTER 11. STATISTICS
(c) What mark must one obtain in order to be in the top 2% of the class?
(d) Approximately 84% of the pupils passed the test. What was the pass mark?
(e) Is the distribution normal or skewed?
3. In a road safety study, the speed of 175 cars was monitored along a specific stretch of highway
in order to find out whether there existed any link between high speed and the large number of
accidents along the route. A frequency table of the results is drawn up below.
Speed (km.h 1 )
Number of cars (Frequency)
50
19
60
28
70
23
80
56
90
20
100
16
110
8
120
5
The mean speed was determined to be around 82 kmhr 1 whi le the median speed was worked
out to be around 84,5 kmhr 1 .
(a) Draw a frequency polygon to visualise the data in the table above.
(b) Is this distribution symmetrical or skewed left or right? Give a reason fro your answer.
f/Vj More practice Crj video solutions (9 J or help at www.everythingmaths.co.za
(1.)01aw (2.) 01 ax (3.) 01 ay
11.3 Extracting a Sample
Population
Suppose you are trying to find out what percentage of South Africa's population owns a car. One way
of doing this might be to send questionnaires to peoples homes, asking them whether they own a car.
However, you quickly run into a problem: you cannot hope to send every person in the country a
questionnaire, it would be far to expensive. Also, not everyone would reply. The best you can do is
send it to a few people, see what percentage of these own a car, and then use this to estimate what
percentage of the entire country own cars. This smaller group of people is called the sample popula
tion.
The sample population must be carefully chosen, in order to avoid biased results. How do we do
this?
First, it must be representative. If all of our sample population comes from a very rich area, then almost
all will have cars. But we obviously cannot conclude from this that almost everyone in the country has
a car! We need to send the questionnaire to rich as well as poor people.
Secondly, the size of the sample population must be large enough. It is no good having a sample
population consisting of only two people, for example. Both may very well not have cars. But we
obviously cannot conclude that no one in the country has a car! The larger the sample population
size, the more likely it is that the statistics of our sample population corresponds to the statistics of the
entire population.
166
CHAPTER 11. STATISTICS
11.4
So how does one ensure that ones sample is representative? There are a variety of methods avail
able, which we will look at now.
Random Sampling. Every person in the country has an equal chance of being selected. It is
unbiased and also independent, which means that the selection of one person has no effect on
the selection on another. One way of doing this would be to give each person in the country a
number, and then ask a computer to give us a list of random numbers. We could then send the
questionnaire to the people corresponding to the random numbers.
Systematic Sampling. Again give every person in the country a number, and then, for example,
select every hundredth person on the list. So person with number 1 would be selected, person
with number 100 would be selected, person with number 200 would be selected, etc.
Stratified Sampling. We consider different subgroups of the population, and take random sam
ples from these. For example, we can divide the population into male and female, different ages,
or into different income ranges.
Cluster Sampling. Here the sample is concentrated in one area. For example, we consider all
the people living in one urban area.
Exercise 112
1 . Discuss the advantages, disadvantages and possible bias when using
(a) systematic sampling
(b) random sampling
(c) cluster sampling
2. Suggest a suitable sampling method that could be used to obtain information on:
(a) passengers views on availability of a local taxi service.
(b) views of learners on school meals.
(c) defects in an item made in a factory.
(d) medical costs of employees in a large company.
3. 2% of a certain magazines' subscribers is randomly selected. The random number 16 out of 50,
is selected. Then subscribers with numbers 16; 66; 116; 166; ... are chosen as a sample. What
kind of sampling is this?
More practice CrJ video solutions C'?) or help at www.everythingmaths
(l.)Olaz (2.)01b0 (3.) Olbl
7 1 .4 Function Fitting and
Regression Analysis
In Grade 1 1 we recorded two sets of data (bivariate data) on a scatter plot and then we drew a line of
best fit as close to as many of the data items as possible. Regression analysis is a method of finding
107
7 7.4
CHAPTER 11. STATISTICS
out exactly which function best fits a given set of data. We can find out the equation of the regression
line by drawing and estimating, or by using an algebraic method called "the least squares method",
available on most scientific calculators. The linear regression equation is written y = a + bx (we say
yhat) or y = A + Bx. Of course these are both variations of a more familiar equation y = mx + c.
Suppose you are doing an experiment with washing dishes. You count how many dishes you begin
with, and then find out how long it takes to finish washing them. So you plot the data on a graph of
time taken versus number of dishes. This is plotted below.
* d
2 3 4 5
Number of dishes
If t is the time taken, and d the number of dishes, then it looks as though t is proportional to d, ie.
t = m .d, where m is the constant of proportionality. There are two questions that interest us now.
1 . How do we find m? One way you have already learnt, is to draw a line of bestfit through the
data points, and then measure the gradient of the line. But this is not terribly precise. Is there a
better way of doing it?
2. How well does our line of best fit really fit our data? If the points on our plot don't all lie close to
the line of best fit, but are scattered everywhere, then the fit is not "good", and our assumption
that t = m .d might be incorrect. Can we find a quantitative measure of how well our line really
fits the data?
In this chapter, we answer both of these questions, using the techniques of regression analysis. ® See
simulation: VMibv at www.everythingmaths.co.za)
Example 1: Fitting by hand
QUESTION
Use
the I
X
the da
ine thi
1,0
ta give
\t best
2,4
n to d
seems
3,1
r aw a .
to fit
4,9
catter
hedat
5,6
plot and line of best fit. Now write down the equation of
a.
6,2
y
2,5
2,8
3,0
4,8
5,1
5,3
k;n
CHAPTER 11. STATISTICS
11.4
SOLUTION
Step 1 : Drawing the graph
The first step is to draw the graph. This is shown below.
V
7 
6 
^
5 
• ^*^"^
4 
3 
^^
• ^^""^ 1
2 
^^ \
1 
n
1 1 1 1 1 1  T
U   II'
12 3 4 5 6
Step 2 : Calculating the equation of the line
The equation of the line is
y = rax + c
From the graph we have drawn, we estimate the yintercept to be 1,5. We estimate
that y = 3,5 when x = 3. So we have that points (3; 3,5) and (0: 1,5) lie on the
line. The gradient of the line, m, is given by
2/2 y\
m =
x 2 — Xi
3,51,5
30
2
3
So we finally have that the equation of the line of best fit is
2
y = 3 X +
The Method of Least Squares
EMCCY
We now come to a more accurate method of finding the line of bestfit. The method is very simple.
Suppose we guess a line of bestfit. Then at at every data point, we find the distance between the
data point and the line. If the line fitted the data perfectly, this distance should be zero for all the data
points. The worse the fit, the larger the differences. We then square each of these distances, and add
l(i<)
7 7.4
CHAPTER 11. STATISTICS
them all together.
The bestfit line is then the line that minimises the sum of the squared distances.
Suppose we have a data set of n points {(#1; 3/1), (x 2 ; 2/2), . . . , (x„; y n )}. We also have a line f(x) =
mx + c that we are trying to fit to the data. The distance between the first data point and the line, for
example, is
distance = y\ — /(si) = y\ — (mx\ + c)
We now square each of these distances and add them together. Lets call this sum S(m,c). Then we
have that
S(m,c)
(Vi  f(xi)f + (V2  f(x 2 )) 2 + .s + (y n  f{x n )f
X>  f {x ^
Thus our problem is to find the value of m and c such that S(m,c) is minimised. Let us call these
minimising values m and c a . Then the line of bestfit is f(x) = m x + Co. We can find m and c
using calculus, but it is tricky, and we will just give you the result, which is that
mo
Ca
i=l xiyi ^LgigLgijij
n T,7=i( x ') 2  (E?=i a; 2
^ n n
1 ^^ m r^
 > yt > Xi=y — m a x
See video: VMhyr at www.everythingmaths.co.za
Example 2: Method of Least Squares
QUESTION
In the table below, we have the records of the maintenance costs in Hands, compared with
the age of the appliance in months. We have data for five appliances.
appliance
1
2
3
4
5
age (x)
5
10
15
20
30
cost (y)
90
140
250
300
380
170
CHAPTER 11. STATISTICS
11.4
SOLUTION
appliance
X
y
xy
x 1
1
5
90
450
25
2
10
140
1400
100
3
15
250
3750
225
4
20
300
6000
400
5
30
380
11400
900
Total
80
1160
23000
1650
:J2xyJ2xJ2y _5x 23000  80 x 1160
nX> 2 Q» 2 5xl650i
1160 12 x 80
12
y — bx
:40
40 + 12a:
Example 3: Using the Sharp EL53 7 VH calculator
QUESTION
Find a regression equation for the following data:
Days (x)
1
2
3
4
5
Growth in m (y)
1,00
2,50
2,75
3,00
3,50
SOLUTION
Step 1 : Getting your calculator ready
Using your calculator, change the mode from normal to "Stat xy". This mode
enables you to type in bivariate data.
Step 2 : Entering the data
Key in the data as follows:
171
7 7.4
CHAPTER 11. STATISTICS
1
(x,y)
1
2,5
2,75
3,0
3,5
DATA
n= 1
2
(x,y)
DATA
n = 2
3
(x,y)
DATA
n = 3
4
{x,y)
DATA
n = 4
5
(x,y)
DATA
n = 5
Step 3 : Getting regression results from the calculator
Ask for the values of the regression coefficients a and b.
RCL
RCL
a gives a = 0,9
6 gives b = 0,55
.■.£ = 0,9 + 0,55x
Example 4: Using the CASIO fx82ES Natural Display calculator
QUESTION
Using a calculator determine the least squares line of best fit for the following data set of
marks.
Learner
1
2
3
4
5
Chemistry (%)
52
55
86
71
45
Accounting (%)
48
64
95
79
50
For a Chemistry mark of 65%, what mark does the least squares line predict for Account
ing^
SOLUTION
Step 1 : Getting your calculator ready
Switch on the calculator. Press [MODE] and then select STAT by pressing [2].
The following screen will appear:
1
lVAR
2
A + BX
3
+CX 2
4
InX
5
eX
6
A.B'X
7
A.XB
8
1/X
Now press [2] for linear regression. Your screen should look something like this:
V
Step 2 : Entering the data
172
CHAPTER 11. STATISTICS
11.4
Press [52] and then [ = ] to enter the first mark under x. Then enter the other
values, in the same way, for the xvariable (the Chemistry marks) in the order
in which they are given in the data set. Then move the cursor across and up
and enter 48 under y opposite 52 in the xcolumn. Continue to enter the other
j/values (the Accounting marks) in order so that they pair off correctly with the
corresponding xvalues.
52
55
y
Then press [AC]. The screen clears but the data remains stored.
1
3
Type
Edit
2:
4:
Data
Sum
5
Var
6:
MinMax
7
Reg
Now press [SHIFT][1] to get the stats computations screen shown below.
Choose Regression by pressing [7].
1:
A
2:
B
3:
r
4:
X
5:
y
Step 3 : Getting regression results from the calculator
a) Press [1] and [ = ] to get the value of the yintercept, a = —5,065 . . . = —5,07 (to two
decimal places)
Finally, to get the slope, use the following key sequence: [SHIFT][1][7][2][ = ]. The calcu
lator gives b = 1,169 . . . = 1,17 (to two decimal places)
The equation of the line of regression is thus:
y = 5,07+ 1,17a;
b) Press [AC][65][SHIFT][1][7][5][ = ]
This gives a (predicted) Accounting mark of"= 70,94 = 71%
Exercise 113
1 . The table below lists the exam results for five students in the subjects of Science and Biology.
Learner
1
2
3
4
5
Science %
55
66
74
92
47
Biology %
48
59
68
81
53
(a) Use the formulae to find the regression equation coefficients a and b.
(b) Draw a scatter plot of the data on graph paper.
(c) Now use algebra to find a more accurate equation.
2. Footlengths and heights of seven students are given in the table below.
Height (cm)
170
163
131
181
146
134
166
Footlength (cm)
27
23
20
28
22
20
24
173
7 7.4
CHAPTER 11. STATISTICS
(a) Draw a scatter plot of the data on graph paper.
(b) Identify and describe any trends shown in the scatter plot.
(c) Find the equation of the least squares line by using algebraic methods and draw the line on
your graph.
(d) Use your equation to predict the height of a student with footlength 21,6 cm.
(e) Use your equation to predict the footlength of a student 176 cm tall.
3. Repeat the data in Question 2 and find the regression line using a calculator.
A" 1 ) More practice f ►) video solutions CfJ or help at www.everythingmaths.co.za
(1.)01b2 (2.)01b3 (3.)01b4
Correlation Coefficients
EMCDA
Once we have applied regression analysis to a set of data, we would like to have a number that tells
us exactly how well the data fits the function. A correlation coefficient, r, is a tool that tells us to what
degree there is a relationship between two sets of data. The correlation coefficient r e [—1; 1] when
r = — 1, there is a perfect negative correlation, when r = 0, there is no correlation and r = 1 is a
perfect positive correlation.
Positive, strong Positive, fairly strong Positive, weak
r sa 0,9 r « 0,7 r ss 0,4
No association Negative, fairly strong
r = rss —0,7
We often use the correlation coefficient r 2 in order to examine the strength of the correlation only.
In this case:
r 2 =0
no correlation
< r 2 < 0,25
very weak
0,25 < r 2 < 0,5
weak
0,5 < r 2 < 0,75
moderate
0,75 < r 2 < 0,9
strong
0,9 < r 2 < 1
very strong
r 2 = l
perfect correlation
The correlation coefficient r can be calculated using the formula
r _ l y^ I !  ■'' \ / ■"  'i
171
CHAPTER 11. STATISTICS
11.4
• where n is the number of data points,
• s x is the standard deviation of the xvalues and
• s y is the standard deviation of the j/values.
This is known as the Pearson's product moment correlation coefficient. It is a long calculation and
much easier to do on the calculator where you simply follow the procedure for the regression equation,
and go on to find r.
Chapter 1 1
End of Chapter Exercises
1. Below is a list of data concerning 12 countries and their respective carbon dioxide
(CO2) emmission levels per person and the gross domestic product (GDP is a measure
of products produced and services delivered within a country in a year) per person.
CO2 emmissions per capita (x)
GDP per capita (y)
South Africa
8,1
3 938
Thailand
2,5
2 712
Italy
7,3
20 943
Australia
17.0
23 893
China
2.5
816
India
0,9
463
Canada
16,0
22 537
United Kingdom
9,0
21785
United States
19,9
31806
Saudi Arabia
11,0
6 853
Iran
3,8
1493
Indonesia
1,2
986
(a) Draw a scatter plot of the data set and your estimate of a line of best fit.
(b) Calculate equation of the line of regression using the method of least squares.
(c) Draw the regression line equation onto the graph.
(d) Calculate the correlation coefficient r.
(e) What conclusion can you reach, regarding the relationship between CO2 emis
sion and GDP per capita for the countries in the data set?
2. A collection of data on the peculiar investigation into a foot size and height of stu
dents was recorded in the table below. Answer the questions to follow.
Length of right foot (cm)
Height (cm)
25,5
163,3
26.1
164,9
23,7
165,5
26,4
173,7
27.5
174,4
24
156
22,6
155,3
27,1
169,3
(a) Draw a scatter plot of the data set and your estimate of a line of best fit.
(b) Calculate equation of the line of regression using the method of least squares or
your calculator.
(c) Draw the regression line equation onto the graph.
175
7 7.4
CHAPTER 11. STATISTICS
(d) Calculate the correlation coefficient r.
(e) What conclusion can you reach, regarding the relationship between the length
of the right foot and height of the students in the data set?
3. A class wrote two tests, and the marks for each were recorded in the table below. Full
marks in the first test was 50, and the second test was out of 30.
(a) Is there a strong association between the marks for the first and second test?
Show why or why not.
(b) One of the learners (in Row 18) did not write the second test. Given her mark
for the first test, calculate an expected mark for the second test.
Learner
Test 1
(Full marks: 50)
Test 2
(Full marks: 30)
1
42
25
2
32
19
3
31
20
4
42
26
5
35
23
6
23
14
7
43
24
8
23
12
9
24
14
10
15
10
11
19
11
12
13
10
13
36
22
14
29
17
15
29
17
16
25
16
17
29
18
18
17
19
30
19
20
28
17
A fast food company produces hamburgers. The number of hamburgers made, and
the costs are recorded over a week.
Hamburgers made
Costs
495
R2 382
550
R2 442
515
R2 484
500
R2 400
480
R2 370
530
R2 448
585
R2 805
(a) Find the linear regression function that best fits the data.
(b) If the total cost in a day is R2 500, estimate the number of hamburgers produced.
(c) What is the cost of 490 hamburgers?
5. The profits of a new shop are recorded over the first 6 months. The owner wants
to predict his future sales. The profits so far have been R90 000; R93 000; R99 500;
R102 000; R101 300; R109 000.
(a) For the profit data, calculate the linear regression function.
(b) Give an estimate of the profits for the next two months.
(c) The owner wants a profit of R130 000. Estimate how many months this will take.
6. A company produces sweets using a machine which runs for a few hours per day.
The number of hours running the machine and the number of sweets produced are
recorded.
170
CHAPTER 11. STATISTICS
11.4
Machine hours
Sweets produced
3,80
275
4,23
287
4,37
291
4,10
281
4,17
286
Find the linear regression equation for the data, and estimate the machine hours
needed to make 300 sweets.
More practice f ►) video solutions C'f) or help at www.everythingmaths.co.:
(1.)01b5 (2.)01b6 (3.)01b7 (4.)01b8 (5.)01b9 (6.)01ba
177
Combinations and
Permutations
1 2. 1 Introduction
Mathematics education began with counting. In the beginning, fingers, beans and buttons were used
to help with counting, but these are only practical for small numbers. What happens when a large
number of items must be counted?
This chapter focuses on how to use mathematical techniques to count combinations of items.
© See introductory video: VMidw at www.everythingmaths.co.za
12.2 Counting
An important aspect of probability theory is the ability to determine the total number of possible
outcomes when multiple events are considered.
For example, what is the total number of possible outcomes when a die is rolled and then a coin is
tossed? The roll of a die has six possible outcomes (1; 2; 3; 4; 5 or 6) and the toss of a coin, 2 outcomes
(head or tails). Counting the possible outcomes can be tedious.
Making a List
EMCDD
The simplest method of counting the total number of outcomes is by making a list:
1H; IT; 2H; 2T; 3H; ST; AH; AT; 5H; 5T; 6H; 6T
or drawing up a table:
die
coin
1
H
1
T
2
H
2
T
3
H
3
T
4
H
4
T
5
H
5
T
6
H
6
T
178
CHAPTER 12. COMBINATIONS AND PERMUTATIONS
12.3
Both these methods result in 12 possible outcomes, but both these methods have a lot of repetition.
Maybe there is a smarter way to write down the result?
Tree Diagrams
EMCDE
One method of eliminating some of the repetition is to use tree diagrams. Tree diagrams are a graphical
method of listing all possible combinations of events from a random experiment.
H T H T H T H T H T H T
Figure 12.1: Example of a tree diagram. Each possible outcome is a branch of the tree.
12.3 Notation
Factorial Notation
EMCDG
For an integer n, the notation n! (read n factorial) represents:
n x (n — 1) x (n — 2) x ■ ■ ■ x 3 x 2 x 1
with the following definition: 0! = 1.
The factorial notation will be used often in this chapter.
12.4 Fundamental Counting
Principle
The use of lists, tables and tree diagrams is only feasible for events with a few outcomes. When the
number of outcomes grows, it is not practical to list the different possibilities and the fundamental
counting principle is used.
The fundamental counting principle describes how to determine the total number of outcomes of a
series of events.
Suppose that two experiments take place. The first experiment has n\ possible outcomes, and the
second has n 2 possible outcomes. Therefore, the first experiment, followed by the second experiment,
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12.5
CHAPTER 12. COMBINATIONS AND PERMUTATIONS
will have a total of rn x n 2 possible outcomes. This idea can be generalised to m experiments as the
total number of outcomes for m experiments is:
n\ X n2 X ri3 X ■ ■ ■ X n„
IL
Yl is the multiplication equivalent of ^2
Note: the order in which the experiments are done does not affect the total number of possible out
comes.
Example 1: Lunch Special
QUESTION
A takeaway has a 4piece lunch special which consists of a sandwich, soup, dessert and drink
for R25,00. They offer the following choices for :
Sandwich: chicken mayonnaise, cheese and tomato, tuna, and ham and lettuce
Soup: tomato, chicken noodle, vegetable
Dessert: icecream, piece of cake
Drink: tea, coffee, coke, Fanta and Sprite.
How many possible meals are there?
SOLUTION
Step 1 : Determine how many parts to the meal there are
There are 4 parts: sandwich, soup, dessert and drink.
Step 2 : Identify how many choices there are for each part
Meal component
Sandwich
Soup
Dessert
Drink
Number of choices
4
3
2
5
Step 3 : Use the fundamental counting principle to determine how many different meals
are possible
4x3x2x5 = 120
So there are 120 possible meals.
12.5 Combinations
The fundamental counting principle describes how to calculate the total number of outcomes when
multiple independent events are performed together.
1N0
CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.5
A more complex problem is determining how many combinations there are of selecting a group of
objects from a set. Mathematically, a combination is defined as an unordered collection of unique
elements, or more formally, a subset of a set. For example, suppose you have fiftytwo playing cards,
and select five cards. The five cards would form a combination and would be a subset of the set of 52
cards.
In a set, the order of the elements in the set does not matter. These are represented usually with curly
braces. For example {2; 4; 6} is a subset of the set {1; 2; 3; 4; 5; 6}. Since the order of the elements
does not matter, only the specific elements are of interest. Therefore,
{2; 4; 6} = {6; 4; 2}
and {1; 1; 1} is the same as {1} because in a set the elements don't usually appear more than once.
So in summary we can say the following: Given S, the set of all possible unique elements, a combina
tion is a subset of the elements of S. The order of the elements in a combination is not important (two
lists with the same elements in different orders are considered to be the same combination). Also, the
elements cannot be repeated in a combination (every element appears once).
Calculating the number of ways that certain patterns can be formed is the beginning of combinatorics,
the study of combinations. Let S be a set with n objects. Combinations of r objects from this set S are
subsets of S having r elements each (where the order of listing the elements does not distinguish two
subsets).
Combination Without Repetition
When the order does not matter, but each object can be chosen only once, the number of combinations
is:
r\(n — r)\ \r
where n is the number of objects from which you can choose and r is the number to be chosen.
For example, if you have 10 numbers and wish to choose 5 you would have 10!/(5!(10 — 5)!) = 252
ways to choose.
For example how many possible 5 card hands are there in a deck of cards with 52 cards?
52!/(5!(52  5)!) = 2 598 960 combinations
Combination with Repetition
When the order does not matter and an object can be chosen more than once, then the number of
combinations is:
(ra + r — 1)! In + r — l\ ln + r — 1
r\(n — 1)! I r J In — 1
where n is the number of objects from which you can choose and r is the number to be chosen.
For example, if you have ten types of donuts to choose from and you want three donuts there are
(10 + 3 — 1)!/3!(10 — 1)! = 220 ways to choose. See video: VMihd at www.everythingmaths.co.za
1S1
72.6
CHAPTER 12. COMBINATIONS AND PERMUTATIONS
Combinatorics and Probability
EMCDK
Combinatorics is quite useful in the computation of probabilities of events, as it can be used to deter
mine exactly how many outcomes are possible in a given experiment.
Example 2: Probability
QUESTION
At a school, learners each play 2 sports. They can choose from netball, basketball, soccer,
athletics, swimming, or tennis. What is the probability that a learner plays soccer and either
netball, basketball or tennis?
SOLUTION
Step 1 : Identify what events we are counting
We count the events: soccer and netball, soccer and basketball, soccer and
tennis. This gives three choices.
Step 2 : Calculate the total number of choices
There are 6 sports to choose from and we choose 2 sports. There are
Q =6!/(2!(62)!) = 15 choices.
Step 3 : Calculate the probability
The probability is the number of events we are counting, divided by the total
number of choices.
Probability = ± =  = 0,2
12.6 Permutations
The concept of a combination did not consider the order of the elements of the subset to be important.
A permutation is a combination with the order of a selection from a group being important. For
example, for the set {1, 2, 3, 4, 5, 6}, the combination {1,2,3} would be identical to the combination
{3, 2, 1}, but these two combinations are different permutations, because the elements in the set are
ordered differently.
More formally, a permutation is an ordered list without repetitions, perhaps missing some elements.
This means that {1; 2; 2; 3; 4; 5; 6} and {1; 2; 4; 5; 5; 6} are not permutations of the set {1; 2; 3; 4; 5; 6}.
182
CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.6
Now suppose you have these objects:
1;2;3
Here is a list of all permutations of all three objects:
12 3; 13 2; 2 13; 2 3 1; 3 12; 3 2 1.
Counting Permutations wemcdm
Let 5 be a set with n objects. Permutations of r objects from this set S refer to sequences of r different
elements of S (where two sequences are considered different if they contain the same elements but
in a different order). Formulae for the number of permutations and combinations are readily available
and important throughout combinatorics.
It is easy to count the number of permutations of size r when chosen from a set of size n (with r < n).
1 . Select the first member of the permutation out of n choices, because there are n distinct elements
in the set.
2. Next, since one of the n elements has already been used, the second member of the permutation
has (n — 1) elements to choose from the remaining set.
3. The third member of the permutation can be filled in (n — 2) ways since 2 have been used
already.
4. This pattern continues until there are r members on the permutation. This means that the last
member can be filled in (n — (r — 1)) = (n — r + 1) ways.
5. Summarising, we find that there is a total of
n(n  l)(n  2) •• • (n — r + 1)
different permutations of r objects, taken from a pool of n objects. This number is denoted by
P(n, r) and can be written in factorial notation as:
P(n,r)
(n — r)\
For example, if we have a total of 5 elements, the integers {1; 2; 3; 4; 5}, how many ways are there for
a permutation of three elements to be selected from this set? In this case, n = 5 and r = 3. Then,
P(5,3) = 5!/7! = 60!.
© See video: VMihe at www.everythingmaths.co.za
Example 3: Permutations
.1 N.>,
12.7
CHAPTER 12. COMBINATIONS AND PERMUTATIONS
QUESTION
Show that a collection of n objects has n\ permutations.
SOLUTION
Proof: Constructing an ordered sequence of n objects is equivalent to choosing the position
occupied by the first object, then choosing the position of the second object, and so on, until
we have chosen the position of each of our n objects.
There are n ways to choose a position for the first object. Once its position is fixed, we
can choose from (n — 1) possible positions for the second object. With the first two placed,
there are (n — 2) remaining possible positions for the third object; and so on. There are only
two positions to choose from for the penultimate object, and the nth object will occupy the
last remaining position.
Therefore, according to the fundamental counting principle, there are
n(n l)(n2) ■ ■ ■ 2 x 1 = n\
ways of constructing an ordered sequence of n objects.
Permutation with Repetition
When order matters and an object can be chosen more than once then the number of
permutations is:
n
where n is the number of objects from which you can choose and r is the number to be chosen.
For example, if you have the letters A, B, C, and D and you wish to discover the number of ways of
arranging them in three letter patterns (trigrams) you find that there are 4 3 or 64 ways. This is because
for the first slot you can choose any of the four values, for the second slot you can choose any of the
four, and for the final slot you can choose any of the four letters. Multiplying them together gives the
total.
12.7 Applications
Extension:
The Binomial Theorem
In mathematics, the binomial theorem is an important formula giving the expansion of powers
of sums. Its simplest version reads
/ \n \~^ / n \ k nk
(x+ y ) =2^\ k \ x y
k = ' '
181
CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.7
Whenever n is a positive integer, the numbers
k k\(nk)\
are the binomial coefficients (the coefficients in front of the powers).
For example, here are the cases n = 2, n = 3 and n = 4:
(x + %i) = x + Ixy + y
(x + yf = x 3 + 3x 2 y + 3xy 2 + y 3
I i \4 4 . ,. 3 ,«22.^ 3, 4
(a; + y) = x + 4x y + bx y + 4xy + j/
The coefficients form a triangle, where each number
is the sum of the two numbers above it:
This formula and the triangular arrangement of the binomial coefficients, are often at
tributed to Blaise Pascal who described them in the 17th century. It was, however, known
to the Chinese mathematician Yang Hui in the 1 3th century, the earlier Persian mathematician
Omar Khayym in the 1 1th century, and the even earlier Indian mathematician Pingala in the
3rd century BC.
Example 4: Number Plates
QUESTION
The number plate on a car consists of any 3 letters of the alphabet (excluding the vowels and
'Q'), followed by any 3 digits (0 to 9). For a car chosen at random, what is the probability that
the number plate starts with a 'Y' and ends with an odd digit?
SOLUTION
Step 1 : Identify what events are counted
The number plate starts with a 'Y', so there is only 1 choice for the first letter,
and ends with an odd digit, so there are 5 choices for the last digit (1, 3, 5, 7, 9).
Step 2 : Find the number of events
Use the counting principle. For each of the other letters, there are 20 possible
choices (26 in the alphabet, minus 5 vowels and 'Q') and 10 possible choices for
each of the other digits.
Number of events = 1 x 20 x 20 x 10 x 10 x 5 = 200 000
Step 3 : Find the number of total number of possible number plates
Use the counting principle. This time, the first letter and last digit can be any
thing.
Total number of choices = 20 x 20 x 20 x 10 x 10 x 10 = 8 000 000
Step 4 : Calculate the probability
The probability is the number of events we are counting, divided by the total
number of choices.
185
12.7
CHAPTER 12. COMBINATIONS AND PERMUTATIONS
Probability  ^§ = £ = 0,025
Example 5: Factorial
QUESTION
Show that
SOLUTION
(nl)\
Method 1: Expand the factorial notation.
n! _ n x (n — 1) x (n — 2) x ■ ■ ■ x 2 x 1
(n1)! ~ (n 1) x (n2) x ■■■ x 2 x 1
Cancelling the common factor of (n — 1) x (n — 2) x ■ ■ ■ x 2 x 1 on the top and bottom leaves
n.
So T^h = n
Method 2: We know that P(n,r) = , re " ! ,, is the number of permutations of r objects,
taken from a pool of n objects. In this case, r = 1. To choose 1 object from n objects, there
are n choices.
So
(nl)l
Chapter 12
End of Chapter Exercises
1 . Tshepo and Sally go to a restaurant, where the menu is:
Starter
Main Course
Dessert
Chicken wings
Beef burger
Chocolate ice cream
Mushroom soup
Chicken burger
Strawberry ice cream
Greek salad
Chicken curry
Apple crumble
Lamb curry
Chocolate mousse
Vegetable lasagna
(a) How many different combinations (of starter, main course, and dessert) can
Tshepo have?
(b) Sally doesn't like chicken. How many different combinations can she have?
186
CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.7
2. Four coins are thrown, and the outcomes recorded. How many different ways are
there of getting three heads? First write out the possibilities, and then use the formula
for combinations.
3. The answers in a multiple choice test can be A, B, C, D, or E. In a test of 12 ques
tions, how many different ways are there of answering the test?
4. A girl has 4 dresses, 2 necklaces, and 3 handbags.
(a) How many different choices of outfit (dress, necklace and handbag) does she
have?
(b) She now buys 2 pairs of shoes. How many choices of outfit (dress, necklace,
handbag and shoes) does she now have?
5. In a soccer tournament of 9 teams, every team plays every other team.
(a) How many matches are there in the tournament?
(b) If there are 5 boys' teams and 4 girls' teams, what is the probability that the first
match will be played between 2 girls' teams?
6. The letters of the word "BLUE" are rearranged randomly. How many new words (a
word is any combination of letters) can be made?
7. The letters of the word "CHEMISTRY" are arranged randomly to form a new word.
What is the probability that the word will start and end with a vowel?
8. There are 2 History classes, 5 Accounting classes, and 4 Mathematics classes at
school. Luke wants to do all three subjects. How many possible combinations of
classes are there?
9. A school netball team has 8 members. How many ways are there to choose a captain,
vicecaptain, and reserve?
10. A class has 15 boys and 10 girls. A debating team of 4 boys and 6 girls must be
chosen. How many ways can this be done?
1 1 . A secret pin number is 3 characters long, and can use any digit (0 to 9) or any letter
of the alphabet. Repeated characters are allowed. How many possible combinations
are there?
(A 1 ) More practice \w\ video solutions (?) or help at www.everythingmaths.co.za
(1.) Olbb (2.) 01 be (3.)01bd (4.) 01 be (5.)01bf (6.)01bg
(7.)01bh (8.) Olbi (9.) Olbj (10.)01bk (11.)01bm
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